ML Aggarwal Class 9 Maths Solutions Chapter 09 Logarithms

Exercise 9.1

Question 1. It is given that △ABC ≅ △RPQ. Is it true to say that BC = QR? Why?
Answer: When △ABC ≅ △RPQ, the vertices match in the order given - A corresponds to R, B corresponds to P, and C corresponds to Q. This means BC matches with PQ as corresponding sides. Since corresponding sides of congruent triangles are equal, we have BC = PQ. Therefore, BC ≠ QR because QR is not the corresponding side to BC.
In simple words: When two triangles are congruent, matching sides are equal. BC matches with PQ, not QR, so BC cannot equal QR.

Exam Tip: Always check the order of vertex labeling in the congruence statement - the position of each letter determines which sides and angles correspond to each other.

Question 2. "If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent." Is the statement true? Why?
Answer: The statement is not true in general. For two triangles to be congruent using the SAS (Side-Angle-Side) rule, the angle must be included between the two equal sides. If the angle is not between the two sides - that is, if it is not an included angle - then the triangles may not be congruent. This is why the SAS rule specifically requires the angle to be included, making it different from just any angle being equal.
In simple words: Two sides and an angle only prove triangles are congruent if the angle is between the two sides. If the angle is somewhere else, the triangles might not match.

Exam Tip: Remember the distinction between included and non-included angles - the word "included" is crucial in the SAS congruence rule and often tested in reasoning-based questions.

Question 3. In the adjoining figure, AB = AC and AP = AQ. Prove that (i) △APC ≅ △AQB (ii) CP = BQ (iii) ∠APC = ∠AQB.
Answer:
(i) Consider triangles APC and AQB. We are given AB = AC and AP = AQ. The angle ∠PAC is common to both triangles. By the SAS congruence rule, △APC ≅ △AQB.
(ii) Since △APC ≅ △AQB, corresponding sides of these congruent triangles must be equal. Therefore, CP = BQ.
(iii) Since △APC ≅ △AQB, corresponding angles of these congruent triangles must be equal. Therefore, ∠APC = ∠AQB.
In simple words: When two triangles have two matching sides and a shared angle between them, the triangles are the same shape and size. This means all their other sides and angles match too.

Exam Tip: For multi-part proofs, establish the congruence first in part (i), then use that result to derive all subsequent equalities - examiners expect you to reference the congruence statement you proved.

Question 4. In the adjoining figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. Prove that (i) △APC ≅ △AQB (ii) CP = BQ (iii) ∠ACP = ∠ABQ.
Answer:
(i) In triangles APC and AQB, we have AP = AQ (given) and AB = AC (given). The angles ∠PAC and ∠QAB are vertically opposite at vertex A, so ∠PAC = ∠QAB. By the SAS rule, △APC ≅ △AQB.
(ii) From the congruence △APC ≅ △AQB, the corresponding sides are equal. Therefore, CP = BQ.
(iii) From the congruence △APC ≅ △AQB, the corresponding angles are equal. Therefore, ∠ACP = ∠ABQ.
In simple words: Two triangles are congruent when they have two equal sides and the same angle at their shared vertex. Once congruent, all matching sides and angles become equal.

Exam Tip: Be careful about identifying vertically opposite angles - these are angles formed at the intersection point and are always equal without needing to be proved.

Question 5. In the adjoining figure, AD = BC and BD = AC. Prove that ∠ADB = ∠BCA and ∠DAB = ∠CBA.
Answer: In triangles ABD and BAC, we have AD = BC (given) and BD = AC (given). The side AB is common to both triangles. By the SSS (Side-Side-Side) congruence rule, △ABD ≅ △BAC. Since the triangles are congruent, their corresponding angles must be equal. Therefore, ∠ADB = ∠BCA and ∠DAB = ∠CBA.
In simple words: When all three sides of one triangle match the three sides of another triangle, the triangles are identical. This means their angles match too.

Exam Tip: The SSS rule is the strongest congruence criterion because it requires checking all three pairs of sides - when all three match, congruence is guaranteed.

Question 6. In the adjoining figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that (i) △ABD ≅ △BAC (ii) BD = AC (iii) ∠ABD = ∠BAC.
Answer:
(i) In triangles ABD and BAC, we are given AD = BC and ∠BAD = ∠ABC. The side AB is shared by both triangles. By the SAS rule, △ABD ≅ △BAC.
(ii) Since △ABD ≅ △BAC, corresponding sides are equal. Therefore, BD = AC.
(iii) Since △ABD ≅ △BAC, corresponding angles are equal. Therefore, ∠ABD = ∠BAC.
In simple words: When two triangles have one equal side, another equal side, and an equal angle between them, they are congruent - meaning all other sides and angles match.

Exam Tip: In quadrilateral problems, drawing both possible triangles and showing their congruence often proves multiple equalities at once - this is an efficient approach.

Question 7. In the adjoining figure, AB = DC and AB || DC. Prove that AD = BC.
Answer: In triangles ABD and CDB, we have AB = DC (given). Since AB is parallel to DC, the alternate angles formed by the transversal BD are equal, so ∠ABD = ∠CDB. The side BD is common to both triangles. By the SAS rule, △ABD ≅ △CDB. Since the triangles are congruent, their corresponding sides are equal. Therefore, AD = BC.
In simple words: When two sides of a quadrilateral are equal and parallel, the other two sides must also be equal. This happens because the diagonals create matching triangles.

Exam Tip: When a line is parallel to another, alternate angles with any transversal cutting both lines are always equal - use this property whenever you see parallel lines in a figure.

Question 8. In the adjoining figure AC = AE, AB = AD and ∠BAD = ∠CAE. Show that BC = DE.
Answer: Given ∠BAD = ∠CAE, add ∠DAC to both sides: ∠BAD + ∠DAC = ∠CAE + ∠DAC. This gives us ∠BAC = ∠DAE. Now consider triangles ABC and ADE. We have AC = AE (given), AB = AD (given), and ∠BAC = ∠DAE (proved above). By the SAS rule, △ABC ≅ △ADE. Therefore, their corresponding sides are equal, and BC = DE.
In simple words: If you add the same angle to both sides of an equal angle, you get two new angles that are equal. Using this fact along with equal sides proves the triangles are the same.

Exam Tip: When angles are not directly equal but involve a common angle, add or subtract that common angle from both given angles to find the relationship you need.

Question 9. In the adjoining figure, AB = AC and D is the midpoint of BC. Use SSS rule of congruency to show that (i) △ABD ≅ △ACD (ii) AD is bisector of ∠A (iii) AD is perpendicular to BC.
Answer:
(i) In triangles ABD and ACD, we have AB = AC (given), and since D is the midpoint of BC, we have BD = CD. The side AD is common to both triangles. By the SSS rule, △ABD ≅ △ACD.
(ii) Since △ABD ≅ △ACD, corresponding angles are equal. Therefore, ∠BAD = ∠CAD, which means AD bisects angle A.
(iii) Since △ABD ≅ △ACD, corresponding angles are equal, so ∠ADB = ∠ADC. These two angles form a straight line along BC, so ∠ADB + ∠ADC = 180°. Since ∠ADB = ∠ADC, we have 2∠ADB = 180°, giving ∠ADB = 90°. Therefore, AD is perpendicular to BC.
In simple words: In an isosceles triangle where two sides are equal, the line from the top vertex to the midpoint of the opposite side has three special properties - it divides the triangle equally, splits the angle in half, and meets the base at a right angle.

Exam Tip: This is a classic result about isosceles triangles - remember that the median from the apex, the angle bisector, and the perpendicular are all the same line in an isosceles triangle.

Question 10. In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that (i) △ACE ≅ △DBF (ii) AE = DF.
Answer:
(i) From the given condition AB = CD, we can add BC to both sides: AB + BC = CD + BC. This simplifies to AC = BD. Now in triangles ACE and DBF, we have AC = BD (just derived), ∠ACE = ∠DBF (given), and CE = BF (given). By the SAS rule, △ACE ≅ △DBF.
(ii) Since △ACE ≅ △DBF, corresponding sides are equal. Therefore, AE = DF.
In simple words: To show two triangles are congruent, you sometimes need to use the given information to find a hidden equal side or angle, then apply a congruence rule.

Exam Tip: When the immediate sides don't look equal, try adding or subtracting common segments from both sides of an equation - this often reveals the third condition needed for a congruence proof.

Question 11. Two line segments AC and BD bisect each other at P. Draw the diagram and prove that (i) AB = CD (ii) ∠BAC = ∠DCA.
Answer:
(i) Since P bisects both AC and BD, we have PA = PC and PB = PD. In triangles BPA and CPD, we have PA = PC, PB = PD, and ∠BPA = ∠CPD (vertically opposite angles). By the SAS rule, △BPA ≅ △CPD. Therefore, corresponding sides are equal, and AB = CD.
(ii) From the congruence △BPA ≅ △CPD, we have ∠DCP = ∠PAB (corresponding angles). From the figure, ∠DCP = ∠DCA and ∠PAB = ∠BAC. Therefore, ∠BAC = ∠DCA.
In simple words: When two line segments cross and cut each other in half, the four triangles formed at the corners are arranged in pairs that match perfectly. This makes opposite sides equal and opposite angles equal.

Exam Tip: Bisecting line segments create vertically opposite angles at the intersection - always use this fact to establish one of the congruence conditions.

Question 12. Prove that the median drawn from the vertex P of an isosceles triangle △PQR with PQ = PR is perpendicular to QR and bisects ∠P.
Answer: Let PM be the median from P to the midpoint M of QR in triangle PQR where PQ = PR. In triangles PQM and PRM, we have PQ = PR (given), QM = RM (since M is the midpoint of QR), and PM is common to both triangles. By the SSS rule, △PQM ≅ △PRM. From this congruence, we know ∠QPM = ∠RPM, which means PM bisects angle P. Also, ∠PMQ = ∠PMR (corresponding angles from congruent triangles). Since these angles form a straight line, ∠PMQ + ∠PMR = 180°, which gives ∠PMQ = ∠PMR = 90°. Therefore, PM is perpendicular to QR.
In simple words: In an isosceles triangle, the line from the tip to the middle of the bottom side does three things - it cuts the angle at the tip in half, it stands straight up at the bottom, and it divides the triangle into two matching pieces.

Exam Tip: This theorem combines the properties of medians, angle bisectors, and perpendiculars in isosceles triangles - many questions build on this result, so memorize all three conclusions.

Question 13. In the adjoining figure, find the values of x and y.
Answer: In triangles ABD and CBD, we have AB = BC (given), AD = CD (given), and BD = BD (common side). By the SSS criterion, triangle ABD is congruent to triangle CBD.

Since corresponding angles in congruent triangles are equal:

(y + 5)° = 46° and (2x + 5)° = 35°

From the first equation: y = 41
From the second equation: 2x = 30, so x = 15

In simple words: When two triangles are congruent, their matching angles must be equal. Match up the angles shown, solve for x and y.

Exam Tip: Always identify corresponding angles carefully when triangles are congruent. Use angle matching to set up equations and solve for unknowns.

Exercise 9.2

Question 1. In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of △PQR should be equal to side AB of △ABC so that the two triangles are congruent? Give reason for your answer.
Answer: Given that ∠A = ∠Q and ∠B = ∠R in the two triangles, side AB must equal side QR. When this condition is met, triangle ABC becomes congruent to triangle QRP by the ASA (Angle-Side-Angle) criterion, since the side connects the two given equal angles.
In simple words: The matching side must connect the two angles that are already equal. In this case, AB connects angles A and B, so it must match with QR, which connects angles Q and R.

Exam Tip: For ASA congruence, the side must be between the two angles. Always verify that the side you choose lies between the two given equal angles.

Question 2. In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of △PQR should be equal to side BC of △ABC so that the two triangles are congruent? Give reason for your answer.
Answer: Given that ∠A = ∠Q and ∠B = ∠R, side BC must equal side RP. This makes triangle ABC congruent to triangle QRP by the AAS (Angle-Angle-Side) criterion. Here, BC is a side not between the two equal angles, but the congruence still holds when matched with the corresponding side RP.
In simple words: When using AAS, the side does not have to be between the angles. Just match the side from one triangle to its corresponding side in the other triangle.

Exam Tip: Remember the difference between ASA and AAS — in ASA the side is between the angles, in AAS it is not. Choose your matching side accordingly.

Question 3. "If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent". Is the statement true? Why?
Answer: The statement is false. When two angles and a side match between two triangles, the triangles are congruent only if the side and angles satisfy a specific arrangement. The side must either lie between the two angles (ASA) or be opposite one of them (AAS). If the side is not in one of these positions relative to the angles, the triangles need not be congruent. In other words, position matters — any two angles and any side do not guarantee congruence.
In simple words: Just having two equal angles and one equal side is not enough. The side must be in the right place relative to those angles.

Exam Tip: Always check the position of the given side with respect to the angles. Only ASA and AAS guarantee congruence; random placement of a side with two angles does not.

Question 4. In the adjoining figure, AD is median of △ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BM = CN.
Answer: In triangles BMD and CND, we observe that BD = CD (because AD is a median, so D divides BC into two equal parts), ∠BMD = ∠CND (both measure 90° since BM and CN are perpendicular), and ∠BDM = ∠CDN (these are vertically opposite angles). By the AAS criterion, triangle BMD is congruent to triangle CND. Since corresponding sides of congruent triangles are equal, we have BM = CN.
In simple words: The two triangles BMD and CND match by AAS. When triangles match, their sides must also match, so BM equals CN.

Exam Tip: Identify the median first — it tells you that D is the midpoint of the base, making BD = CD. Use this along with the right angles to establish congruence.

Question 5. In the adjoining figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD.
Answer: In triangles BEM and DEN, we have BM = DN (given), ∠BME = ∠DNE (both equal 90° as BM and DN are perpendiculars), and ∠BEM = ∠DEN (these are vertically opposite angles at point E). By the AAS criterion, triangle BEM is congruent to triangle DEN. From this congruence, corresponding sides give us DE = BE. Since these two segments are equal, point E must be the midpoint of BD, meaning AC bisects BD at E.
In simple words: The two right triangles are congruent, so their matching parts are equal. Equal segments BE and DE show that E is the middle point of BD.

Exam Tip: When a line bisects a segment, it cuts it into two equal parts. Always verify this by showing the two parts are equal using congruent triangles.

Question 6. In the adjoining figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that △ABC ≅ △CDA.
Answer: In triangles ABC and CDA, side AC is common to both. Since l and m are parallel and cut by the same transversal, ∠ACB and ∠CAD are alternate angles, so they are equal. Similarly, since p and q are parallel and cut by the same transversal, ∠BAC and ∠ACD are alternate angles, so they are equal. By the ASA criterion (two angles and the included side), triangle ABC is congruent to triangle CDA.
In simple words: Parallel lines create equal alternate angles. When two angles and the side between them match, the triangles are congruent.

Exam Tip: Look for parallel lines in the figure — they create alternate angles that are equal. Use these equal angles along with a common side to apply ASA.

Question 7. In the adjoining figure, two lines AB and CD intersect each other at the point O such that BC ∥ DA and BC = DA. Show that O is the mid-point of both the line segments AB and CD.
Answer: In triangles BOC and DOA, we have BC = DA (given), ∠CBO = ∠DAO (these are alternate angles since BC is parallel to DA), and ∠BOC = ∠DOA (these are vertically opposite angles). By the ASA criterion, triangle BOC is congruent to triangle DOA. From this congruence, corresponding sides give us BO = AO and CO = DO. This means O divides both AB and CD into two equal segments, so O is the midpoint of both line segments.
In simple words: Two triangles match using ASA. Their equal sides show that O cuts both line segments exactly in half.

Exam Tip: When two intersecting lines meet at a point, check if that point divides both lines equally by proving congruent triangles on either side.

Question 8. In the adjoining figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that (i) △ACD ≅ △BDC, (ii) BC = AD, (iii) ∠A = ∠B
Answer:
(i) In triangles ACD and BDC, we are given ∠BCD = ∠ADC and ∠BCA = ∠ADB. Adding these equations: ∠BCD + ∠BCA = ∠ADC + ∠ADB, which gives ∠ACD = ∠BDC. Side CD is common to both triangles. We also have ∠ADC = ∠BCD (given). By the ASA criterion (two angles and the included side), triangle ACD is congruent to triangle BDC.

(ii) Since triangle ACD is congruent to triangle BDC (from part i), corresponding sides are equal. Therefore, BC = AD.

(iii) Since triangle ACD is congruent to triangle BDC, corresponding angles are equal. Therefore, ∠A = ∠B.
In simple words: Start by proving one pair of triangles are the same using ASA. Then use that match to find equal sides and angles.

Exam Tip: Multi-part proofs build on each other. Prove part (i) first, then use that result for parts (ii) and (iii). Always state which corresponding parts you are using from the congruence.

Question 9. In the adjoining figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. Prove that (i) △EBC ≅ △DCB, (ii) △OEB ≅ △ODC, (iii) OB = OC.
Answer:
(i) Given that ∠ABC = ∠ACB, we can deduce ∠EBC = ∠DCB. In triangles EBC and DCB, we have ∠EBC = ∠DCB (proved), BE = CD (given), and BC = BC (common side). By the SAS criterion, triangle EBC is congruent to triangle DCB.

(ii) From part (i), triangle EBC is congruent to triangle DCB. When we subtract the common triangle OBC from both sides, the remaining regions triangle OEB and triangle ODC must also be congruent. Therefore, triangle OEB is congruent to triangle ODC.

(iii) Since triangle OEB is congruent to triangle ODC, corresponding sides are equal. Therefore, OB = OC.
In simple words: Equal angles in the base give equal angles for the sides. Two triangles match by SAS. Subtracting a common part keeps the remaining parts congruent.

Exam Tip: For part (ii), subtracting common triangular regions is a valid technique to prove congruence of the remaining parts. State the subtraction clearly.

Question 10. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Answer: In triangles APB and APC, we have AB = AC (given as the triangle is isosceles), ∠APB = ∠APC (both measure 90° since AP is perpendicular to BC), and AP = AP (common side). By the RHS (Right angle - Hypotenuse - Side) criterion, triangle APB is congruent to triangle APC. Since corresponding angles of congruent triangles are equal, we have ∠B = ∠C. This demonstrates that in an isosceles triangle, the base angles are equal.
In simple words: Draw a line straight down from the top. Two right triangles form on either side. They are the same shape and size, so their base angles must match.

Exam Tip: In an isosceles triangle, always consider drawing a perpendicular from the vertex angle to the base. This creates two congruent right triangles and helps prove that base angles are equal.

Question 11. In the adjoining figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that △ABC ≅ △DEF.
Answer: Since BF = EC (given), we can add FC to both sides to get BF + FC = EC + FC, which simplifies to BC = EF. Now, in triangles ABC and DEF, we have BA = DE (given), ∠BAC = ∠EDF (both are right angles, as BA ⊥ AC and DE ⊥ DF), and BC = EF (shown above). By the RHS criterion, triangle ABC is congruent to triangle DEF, since we have matching right angles, equal hypotenuses, and an equal side adjacent to the right angle.
In simple words: Use the given equal segments to find that the hypotenuses are equal. With a right angle, equal hypotenuses, and another matching side, the triangles must be the same.

Exam Tip: For RHS congruence, you need a right angle, equal hypotenuses, and one more equal side. Don't forget to derive equal sides from the given information by adding or subtracting equal quantities.

Question 12. ABCD is a rectangle. X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
Answer: In triangles ABY and ABX, we have AB as a common side. Since ABCD is a rectangle, angles XAB and YBA are both 90 degrees. We are also given that AY = BX. Using the RHS (Right angle - Hypotenuse - Side) axiom, triangle ABY is congruent to triangle ABX. Because corresponding parts of congruent triangles are equal, we get BY = AX and angle BAY = angle ABX.
In simple words: The two triangles have the same hypotenuse, one equal side, and both have right angles, so they match exactly. This means the other parts must also match.

Exam Tip: Always identify which sides and angles are equal before stating the congruence axiom. Mark the right angles clearly in your diagram to make the RHS axiom obvious.

Question 13(a). In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of △PQR. If XS ⊥ QR and XT ⊥ PQ, prove that (i) △XTQ ≅ △XSQ (ii) PX bisects the angle P.
Answer: (i) Given that QX is the angle bisector of angle PQR, we have angle PQX = angle XQS. From the figure, angle XTQ = angle XSQ (both equal to 90 degrees), and XQ = XQ (common side). By the ASA axiom, triangle XTQ is congruent to triangle XSQ.
(ii) Draw a perpendicular from X to PR, calling it XU. In triangles XSR and XUR, we have angle XSR = angle XUR (both 90 degrees), angle XRS = angle XRU (since XR bisects angle PRQ), and XR = XR (common). By AAS axiom, triangle XSR is congruent to triangle XUR, so XU = XS. From part (i), we know XS = XT. Now in triangles XUP and XTP, we have XU = XT, XP = XP (common), and angle XUP = angle XTP (both 90 degrees). By SAS axiom, triangle XUP is congruent to triangle XTP. Therefore, angle XPU = angle XPT, which means PX bisects angle P.
In simple words: When angle bisectors meet a perpendicular, they create matching triangles. Using these matching triangles step by step shows that PX must split angle P into two equal parts.

Exam Tip: This is a multi-step proof - construct the perpendicular XU clearly and track which triangles are congruent at each stage. The final conclusion comes from comparing the angles at P.

Question 13(b). In the figure (2) given below, AB || DC and ∠C = ∠D. Prove that (i) AD = BC (ii) AC = BD.
Answer: (i) Draw perpendiculars from A and B to line CD, meeting it at E and F respectively. Looking at triangles ADE and BCF, we have angle ADE = angle BCF (given), angle AED = angle BFC (both 90 degrees), and AE = BF (perpendicular distances between parallel lines are equal). By the AAS axiom, triangle ADE is congruent to triangle BCF. From this congruence, AD = BC.
(ii) Draw diagonals AC and BD. Now consider triangles ACD and BDC. We have angle ADC = angle BCD (given), AD = BC (proved in part i), and DC = DC (common side). By the SAS axiom, triangle ACD is congruent to triangle BDC. Therefore, AC = BD.
In simple words: The perpendiculars from the two upper vertices create a pair of matching triangles, giving us equal sides. Then using those equal sides with the diagonals creates another pair of matching triangles.

Exam Tip: Drawing the perpendiculars in part (i) is the key step - this lets you use the equal-distance property of parallel lines. Part (ii) builds directly on the result from part (i).

Question 13(c). In the figure (3) given below, BA || DF and CA || EG and BD = EC. Prove that (i) BG = DF (ii) EG = CF.
Answer: Given BD = EC, we can add DE to both sides to get BD + DE = DE + EC, which gives us BE = DC. Now consider triangles BGE and DFC. Since BA || DF, angle GBE = angle FDC (corresponding angles). Since CA || EG, angle GEB = angle FCD (corresponding angles). We have also shown that BE = DC. By the ASA axiom, triangle BGE is congruent to triangle DFC. From this congruence, we get BG = DF and EG = CF.
In simple words: The parallel lines make certain angles equal. Adding the same length to two equal pieces makes new equal pieces. These facts together with our parallel lines prove both required equalities.

Exam Tip: The key insight is to add DE to both sides of BD = EC to establish BE = DC. This intermediate result is essential before comparing the two triangles.

Question 14(i). In the following figure, find the values of x and y.
Answer: In triangles ABC and CDE, angle ACB = angle DCE (vertically opposite angles), angle BAC = angle CED (given), and BC = CD (given). By the AAS axiom, triangle ABC is congruent to triangle CDE. From this congruence, DE = AB, which gives us 2y + 3 = 25, so 2y = 22 and y = 11. Also from the given condition BC = CD, we have 3x - 7 = 32, so 3x = 39 and x = 13.
In simple words: The two triangles match using angle-angle-side, so their other matching sides must be equal. This gives us two simple equations to solve for x and y.

Exam Tip: Vertically opposite angles are always equal - mark them clearly in your diagram. Once you establish triangle congruence, list all corresponding equal sides to find the values of the unknowns.

Question 14(ii). In the following figure, find the values of x and y.
Answer: In triangles ABC and ADE, angle BAC = angle DAE (given), angle BCA = angle EDA (given), and AC = AD (given). By the ASA axiom, triangle ABC is congruent to triangle ADE. From this congruence, BC = DE, giving us x = 2y ... (i). Also, AB = AE, so 2x + 4 = 3y + 8. Substituting x = 2y from equation (i), we get 2(2y) + 4 = 3y + 8, which simplifies to 4y + 4 = 3y + 8. Solving, 4y - 3y = 8 - 4, so y = 4. Therefore, x = 2(4) = 8.
In simple words: The triangles are congruent, so matching sides are equal. This gives us two equations. Substitute one into the other to find both unknowns.

Exam Tip: When you have two unknowns, get one equation from each pair of matching sides. Then use substitution to solve the system - this is faster than trying to solve both equations simultaneously.

Exercise 9.3

Question 1. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Answer: Given AB = AC, we know that the angles opposite these equal sides must be equal. Therefore, angle B = angle C. Let's call each of these angles x degrees. The sum of all angles in a triangle is 180 degrees. So, 90 + x + x = 180, which gives us 90 + 2x = 180. Solving, 2x = 90, so x = 45 degrees. Therefore, angle B = angle C = 45 degrees.
In simple words: In any triangle, equal sides have equal angles opposite them. Since the right angle is 90 degrees, the other two must split the remaining 90 degrees equally.

Exam Tip: Remember the isosceles triangle property: angles opposite equal sides are equal. This immediately tells you that angle B = angle C without needing to work it out.

Question 2. Show that the angles of an equilateral triangle are 60° each.
Answer: In an equilateral triangle, all three sides are equal: AB = BC = CA. When all sides are equal, the angles opposite to those sides are also equal. Therefore, angle C = angle A = angle B. Let each angle be x degrees. The sum of angles in a triangle is 180 degrees, so x + x + x = 180, which gives us 3x = 180. Dividing both sides by 3, we get x = 60 degrees. This proves that each angle of an equilateral triangle is 60 degrees.
In simple words: All sides are equal, so all angles must be equal too. Three equal angles that add up to 180 degrees means each one is 180 divided by 3, which is 60 degrees.

Exam Tip: Always start by naming what makes the triangle equilateral - that all three sides are equal. From there, the equal angles follow as a direct consequence.

Question 3. Show that every equiangular triangle is equilateral.
Answer: Let ABC be a triangle where all three angles are equal: angle A = angle B = angle C. If angle A = angle B, then the sides opposite these equal angles must be equal, so BC = AC. If angle B = angle C, then the sides opposite these equal angles must be equal, so AB = BC. Combining these results, we get AB = BC = AC, which means all three sides are equal. Therefore, the triangle is equilateral.
In simple words: Whenever two angles are equal, the sides opposite them are equal. If all three angles match each other, then all three sides must match too.

Exam Tip: Use the property "sides opposite equal angles are equal" twice - once for each pair of equal angles. This gives you three equal sides, proving the triangle is equilateral.

Question 4(i). In the following figure, find the value of x:
Answer: Since triangle ABC is isosceles with AB = AC, the base angles are equal: angle B = angle C = y degrees. The sum of angles in the triangle is 180 degrees. Therefore, 50 + y + y = 180, giving us 50 + 2y = 180. Solving, 2y = 130, so y = 65 degrees. Since angle ACD is on a straight line with angle ACB, these are supplementary angles: angle ACD + angle ACB = 180. So x + 65 = 180, which gives us x = 115 degrees.
In simple words: The isosceles triangle has two equal base angles. Find these first, then use the fact that angles on a straight line add to 180 to find x.

Exam Tip: Don't try to find x directly from the triangle. First find the base angles using the isosceles property, then recognize that x and angle ACB are supplementary angles on the line BCD.

Question 4(ii). In the following figure, find the value of x:
Answer: Triangle PRS is isosceles with PR = SR. This means the angles at the base are equal, so \( \angle S = \angle RPS = 30° \). Now, \( \angle QPS = \angle QPR + \angle RPS = 52° + 30° = 82° \). Since all angles in a triangle sum to 180°, in triangle PQS we have \( \angle P + \angle Q + \angle S = 180° \). Substituting: \( 82° + x° + 30° = 180° \), which gives \( x° + 112° = 180° \), so \( x° = 68° \).
In simple words: When two sides of a triangle are equal, the angles opposite to them are also equal. Use this fact to find the angles, then add all three angles to get 180°.

Exam Tip: Always check whether a triangle is isosceles by looking for equal side markings, and remember that equal sides give equal opposite angles.

Question 4(iii). In the following figure, find the value of x:
Answer: Triangle BDC is isosceles with BD = DC, so the base angles are equal: \( \angle DBC = \angle DCB = 27° \). Using the exterior angle property, the exterior angle \( \angle ADC = \angle DBC + \angle DCB = 27° + 27° = 54° \). Now triangle ADC is also isosceles with AC = DC, giving \( \angle DAC = \angle ADC = 54° \). The angles in triangle ADC must sum to 180°: \( \angle DAC + \angle ADC + \angle DCA = 180° \), so \( 54° + 54° + x = 180° \), which means \( x + 108° = 180° \), giving \( x = 72° \).
In simple words: Find angles using the isosceles triangle property, then use the exterior angle rule. Finally, use the angle sum to find x.

Exam Tip: The exterior angle theorem (exterior angle = sum of two non-adjacent interior angles) is often key to solving multi-triangle problems.

Question 5(i). In the following figure, find the value of x:
Answer: Triangle ABD is isosceles with BD = AD, so \( \angle DBA = \angle BAD = 48° \). By the exterior angle property, \( \angle ADC = \angle BAD + \angle DBA = 48° + 48° = 96° \). Triangle ADC is isosceles with AD = DC, so \( \angle DAC = \angle DCA = x° \). Using the angle sum in triangle ADC: \( x° + 96° + x° = 180° \), which gives \( 2x° + 96° = 180° \), so \( 2x° = 84° \) and \( x° = 42° \).
In simple words: In an isosceles triangle, the two angles at the base are equal. Use this to find unknown angles, then apply the angle sum property.

Exam Tip: When a problem involves two or more triangles, work systematically from the innermost triangle outward, using equal side markings to identify isosceles triangles first.

Question 5(ii). In the following figure, find the value of x:
Answer: From the diagram, \( \angle ACD = 180° - 130° = 50° \) (angles on a straight line). In triangle ADC with AD = CD, the base angles are equal: \( \angle DAC = \angle ACD = 50° \). So \( \angle ADC = 180° - (50° + 50°) = 80° \). Since angles on a straight line sum to 180°, \( \angle ADB = 180° - 80° = 100° \). In triangle ADB with equal sides AB = DB, we have \( \angle DBA = \angle DAB = x° \). Using the angle sum: \( x° + x° + 100° = 180° \), which gives \( 2x° = 80° \) and \( x° = 40° \).
In simple words: Use straight-line angles to find one angle, then use equal sides to find more angles, and finally solve for x.

Exam Tip: Angles on a straight line always add to 180° - this is a simple rule to apply when you see points lying on the same line.

Question 5(iii). In the following figure, find the value of x:
Answer: In triangle ADC with DC = CA, the equal angles are \( \angle ADC = \angle CAD = a \). The angle sum gives \( a + 56° + a = 180° \), so \( 2a = 124° \) and \( a = 62° \). In triangle ABD with AD = BD, we have \( \angle ABD = \angle DAB = (x - a)° \). Since \( \angle BDA \) is supplementary to \( \angle ADC \) on the line, \( \angle BDA = 180° - a = 180° - 62° = 118° \). Wait, let me recalculate: \( \angle ADB = 180° - a \). In triangle ABD: \( (x - a)° + (180 - a)° + (x - a)° = 180° \), which simplifies to \( 2x° - 3a° + 180° = 180° \), giving \( 2x° = 3a° = 3(62°) = 186° \), so \( x° = 93° \).
In simple words: Use angle relationships to set up expressions. In triangle ADC, two sides are equal, so two angles are equal. Then use this to find the angles in triangle ABD and solve for x.

Exam Tip: When angles are expressed as algebraic expressions (like x - a), substitute the numerical value only after finding the expression's value separately.

Question 6(a). In the figure (1) given below, AB = AD, BC = DC. Find \( \angle ABC \).
Answer: Draw the line BD to create two triangles. In triangle ABD where AB = AD, the base angles are equal: \( \angle ABD = \angle ADB = a \). The angle sum gives \( a + a + 54° = 180° \), so \( 2a = 126° \) and \( a = 63° \). In triangle BCD where BC = DC, the base angles are equal: \( \angle CDB = \angle DBC = b \). The angle sum gives \( b + b + 116° = 180° \), so \( 2b = 64° \) and \( b = 32° \). The angle ABC is the sum of these two base angles: \( \angle ABC = a + b = 63° + 32° = 95° \).
In simple words: Draw an extra line to divide the quadrilateral into two triangles. Use the isosceles property on each triangle to find the angles, then add them together.

Exam Tip: When a problem gives you multiple equal sides, consider drawing an auxiliary line to create simpler triangles you can solve separately.

Question 6(b). In the figure (2) given below, BC = CD. Find \( \angle ACB \).
Answer: From the diagram, the exterior angle at A is 138°, so \( \angle DAC = 180° - 138° = 42° \). The angle \( \angle BDC = 180° - 116° = 64° \). In triangle BCD where BC = CD, the equal angles are \( \angle CBD = \angle BDC = 64° \). So \( \angle DCB = 180° - (64° + 64°) = 52° \). In triangle ADC, we find \( \angle DCA = 180° - (42° + 116°) = 22° \). The complete angle \( \angle ACB = \angle DCB + \angle DCA = 52° + 22° = 74° \).
In simple words: Use straight-line angles and the exterior angle to find the basic angles. Then use equal sides to find more angles. Finally, add the parts together.

Exam Tip: When angles are split into multiple parts, find each part separately using different triangles, then add them at the end.

Question 6(c). In the figure (3) given below, AB || CD and CA = CE. If \( \angle ACE = 74° \) and \( \angle BAE = 15° \), find the values of x and y.
Answer: Since CA = CE, triangle ACE is isosceles with equal base angles: \( \angle CAE = \angle CEA = a \). Using the angle sum: \( a + a + 74° = 180° \), so \( 2a = 106° \) and \( a = 53° \). Since angles CEA and AEB are supplementary on line CE: \( 53° + x = 180° \), giving \( x = 127° \). In triangle AEB, the angle sum is \( \angle EAB + \angle AEB + \angle ABE = 180° \). Substituting: \( 15° + 127° + \angle ABE = 180° \), so \( \angle ABE = 38° \). Since AB || CD, alternate angles are equal: \( y = \angle ABE = 38° \).
In simple words: When two sides are equal, the opposite angles are equal. Use this to find one angle, then use properties of lines and parallel lines to find the rest.

Exam Tip: When parallel lines are involved, look for alternate angles and corresponding angles - they are always equal and provide shortcuts to your answer.

Question 7. In \( \triangle ABC \), AB = AC, \( \angle A = (5x + 20)° \) and each of the base angle is \( \frac{2}{5} \) th of \( \angle A \). Find the measure of \( \angle A \).
Answer: Since the triangle is isosceles with AB = AC, the base angles are equal. Given that each base angle is \( \frac{2}{5} \) of angle A: each base angle = \( \frac{2}{5} \times (5x + 20)° = (2x + 8)° \). The angle sum in the triangle is: \( (5x + 20)° + (2x + 8)° + (2x + 8)° = 180° \). Simplifying: \( 9x + 36 = 180 \), so \( 9x = 144 \) and \( x = 16° \). Therefore, \( \angle A = 5(16) + 20 = 100° \).
In simple words: In an isosceles triangle, the two base angles are equal. Write an equation using this fact and the angle sum rule, then solve for the unknown.

Exam Tip: When a problem relates angles using fractions or formulas, express all angles in the same variable and use the angle sum property to create an equation.

Question 8(a). In the figure (1) given below, ABC is an equilateral triangle. Base BC is produced to E, such that BC = CE. Calculate \( \angle ACE \) and \( \angle AEC \).
Answer: Since ABC is equilateral, all its angles are 60°. The angles \( \angle ACB \) and \( \angle ACE \) lie on a straight line, so \( 60° + \angle ACE = 180° \), giving \( \angle ACE = 120° \). In triangle ACE, since BC = CE and the base angles must be equal (triangle is isosceles): \( \angle CAE = \angle AEC = y \). Using the angle sum: \( y + y + 120° = 180° \), so \( 2y = 60° \) and \( y = 30° \). Therefore, \( \angle ACE = 120° \) and \( \angle AEC = 30° \).
In simple words: An equilateral triangle has all angles equal to 60°. When one side is extended, use the straight-line angle to find the next angle, then use equal sides to find the remaining angles.

Exam Tip: Always start with what you know for certain (like 60° in an equilateral triangle) and use straight-line angles and side properties to build up your solution step by step.

Question 8(b). In the figure (2) given below, prove that ∠BAD : ∠ADB = 3 : 1.
Answer: From the figure, we see that \( \angle ADC = \angle CAD \) since angles opposite equal sides in a triangle are the same. This means \( \angle BAD = \angle BAC + \angle CAD = \angle BAC + \angle ADB \). In triangle ABC, \( \angle BAC = \angle ACB \) because angles opposite equal sides match. We can write \( \angle ACB = 180 - \angle ACD = 180 - [180 - (\angle CAD + \angle ADB)] = \angle CAD + \angle ADB = \angle ADB + \angle ADB = 2\angle ADB \). Since \( \angle BAC = \angle ACB \), we have \( \angle BAC = 2\angle ADB \). Putting this into our earlier equation, \( \angle BAD = 2\angle ADB + \angle ADB = 3\angle ADB \). Therefore, \( \angle BAD : \angle ADB = 3\angle ADB : \angle ADB = 3 : 1 \).
In simple words: By using the property that equal sides have equal opposite angles, we can show that angle BAD is three times angle ADB.

Exam Tip: Identify which angles are equal based on equal sides, then use substitution step-by-step to build the ratio. Write each ratio comparison clearly at the end.

Question 8(c). In the figure (3) given below, AB || CD. Find the values of x, y and z.
Answer: Since AB is parallel to CD, angles BAD and ADC are alternate angles, so \( x = 42° \). In triangle CED, the angles add to 180°, giving us \( 24° + \angle CED + 42° = 180° \), which means \( \angle CED = 114° \). Since angles CEA and CED are supplementary (they form a straight line), \( \angle CEA = 180° - 114° = 66° \). In triangle CEA, since two sides are equal, \( \angle CAE = \angle CEA = 66° \), so \( y = 66° \). Finally, in triangle CEA, the three angles sum to 180°: \( z + 66° + 66° = 180° \), giving \( z = 48° \).
In simple words: Use parallel lines to find x, then use angle sums in triangles to find y and z.

Exam Tip: When lines are parallel, mark alternate angles as equal immediately. Use the angle-sum property of triangles at each step and verify your final answers add correctly.

Question 9. In the adjoining figure, D is the midpoint of BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ABC is an isosceles triangle.
Answer: Looking at triangles BED and CFD, we note that both have right angles (since DE and DF are perpendiculars): \( \angle BED = \angle CFD = 90° \). We are given that \( DE = DF \). Since D is the midpoint of BC, we have \( BD = DC \). By the RHS (Right angle - Hypotenuse - Side) congruence rule, \( \triangle BED \cong \triangle CFD \). From this congruence, the corresponding angles are equal, so \( \angle B = \angle C \). When the base angles of a triangle are equal, the sides opposite to them must be equal, giving us \( AC = AB \). Therefore, ABC is isosceles.
In simple words: If two right triangles have the same perpendicular distance and the same hypotenuse length, they are congruent, which makes the outer triangle isosceles.

Exam Tip: Identify the two right triangles carefully and state which sides are equal. Use RHS congruence explicitly and conclude about the base angles.

Question 10. In the adjoining figure, AD, BE and CF are altitudes of △ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.
Answer: Consider triangles BEC and BFC. The hypotenuse BC is common to both. We are given that \( BE = CF \). Since these are altitudes, they are perpendicular to the sides, so \( \angle CFB = \angle CEB = 90° \). By RHS congruence, \( \triangle BEC \cong \triangle BFC \). Therefore, \( \angle C = \angle B \), which means \( AB = AC \) (sides opposite equal angles). Next, compare triangles CFA and ADC. We have \( AD = CF \) (given), \( \angle ADC = \angle CFA = 90° \) (since these are altitudes), and \( AC = AC \) (common). By RHS congruence, \( \triangle CFA \cong \triangle ADC \). This gives \( \angle A = \angle C \), so \( AB = BC \). Combining both results: \( AB = AC \) and \( AB = BC \), we get \( AB = BC = AC \). Therefore, ABC is equilateral.
In simple words: When all three altitudes are equal, each pair of triangles formed by these altitudes turn out to be congruent, forcing the original triangle to have all sides equal.

Exam Tip: Carefully pair the two triangles at each stage and state the RHS congruence. Write the equal angle conclusions clearly, as these lead directly to the equal sides of the main triangle.

Question 11. In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Show that: (i) △DBC ≅ △ECB (ii) ∠DCB = ∠EBC (iii) OB = OC, where O is the point of intersection of BE and CD.
Answer:
(i) From the figure, we have \( BD = CE \) (given), \( BC = BC \) (common side), and \( \angle DBC = \angle ECB \) because \( AB = AC \) (the base angles of an isosceles triangle are equal). By SAS axiom, \( \triangle DBC \cong \triangle ECB \).
(ii) Since \( \triangle DBC \cong \triangle ECB \), their corresponding angles are equal. Therefore, \( \angle DCB = \angle EBC \).
(iii) From the congruence \( \triangle DBC \cong \triangle ECB \), we get \( \angle BDO = \angle CEO \) (by corresponding parts). Also, \( \angle DOB = \angle EOC \) (vertically opposite angles). We are given \( BD = CE \). By ASA axiom, \( \triangle BOD \cong \triangle EOC \). From this congruence, \( OB = OC \).
In simple words: The isosceles condition forces two triangles to be congruent. This congruence then cascades through the figure, making the intersection point O sit exactly halfway between B and C.

Exam Tip: For part (i), state clearly why the angle at B equals the angle at C. For part (iii), extract the correct angle equalities from the congruence found in part (i) before using ASA.

Question 12. ABC is an isosceles triangle in which AB = AC. P is any point in the interior of △ABC such that ∠ABP = ∠ACP. Prove that (a) BP = CP (b) AP bisects ∠BAC.
Answer:
(a) Since \( AB = AC \), the base angles are equal: \( \angle B = \angle C \). We are given that \( \angle ABP = \angle ACP \). Subtracting the second equation from the first: \( \angle B - \angle ABP = \angle C - \angle ACP \), which gives \( \angle PBC = \angle PCB \). When two angles of a triangle are equal, the sides opposite them are equal, so \( BP = CP \).
(b) We now know that \( BP = CP \) (just proved) and \( AB = AC \) (given). We are also given \( \angle ABP = \angle ACP \). By SAS axiom, \( \triangle ABP \cong \triangle ACP \). Therefore, the corresponding angles are equal: \( \angle PAB = \angle PAC \). This means AP divides angle BAC into two equal parts, so AP bisects \( \angle BAC \).
In simple words: The equal base angles of the isosceles triangle, combined with the equal given angles at B and C, force point P to sit on a special line from A that cuts the top angle exactly in half.

Exam Tip: For part (a), subtract the angle equations to isolate the two angles in triangle PBC. For part (b), list all three pairs of equal elements (two sides and one angle) needed for SAS before stating the congruence.

Question 13. ΔPQR is an isosceles triangle such that PQ = QR. If S is a point on QR produced such that PR = RS and ∠QPS = 63°, find ∠PSQ.
Answer: In triangle PQR, since \( PQ = QR \), the angles opposite these equal sides are equal. Let \( \angle QPR = \angle QRP = x° \). In triangle PRS, since \( PR = RS \), we have \( \angle PSR = \angle SPR = y° \). We are given \( \angle QPS = 63° \). From the figure, \( \angle QPR + \angle SPR = 63° \), so \( x° + y° = 63° \). Using the exterior angle theorem, the angle \( \angle PRQ \) (which is \( x° \)) is an exterior angle to triangle PRS at vertex R, so it equals the sum of the two non-adjacent interior angles: \( x° = y° + y° = 2y° \). Substituting into our first equation: \( 2y° + y° = 63° \), giving \( 3y° = 63° \) and \( y° = 21° \). Therefore, \( \angle PSQ = \angle PSR = 21° \).
In simple words: Use the equal sides in each triangle to set up angle equations. The exterior angle property then connects the two triangles, allowing you to solve for the unknown angle.

Exam Tip: When a point is on a line "produced" (extended), the straight angle property and exterior angle theorem become key. Set up equations using these tools methodically.

Question 14. In the adjoining figure, D and E are points on the side BC of △ABC such that BD = EC and AD = AE. Show that △ABD ≅ △ACE.
Answer: We are given \( AD = AE \). Since angles opposite equal sides are equal, \( \angle ADE = \angle AED \). Subtracting each from 180°: \( 180° - \angle ADE = 180° - \angle AED \), which gives \( \angle ADB = \angle AEC \). We are also given \( BD = EC \). The side AB is common to both triangles ABD and ACE (or we compare the appropriate sides). By SAS axiom, comparing triangle ABD and triangle ACE: we have \( AD = AE \), \( \angle ADB = \angle AEC \) (from above), and \( BD = EC \) (given). Therefore, \( \triangle ABD \cong \triangle ACE \).
In simple words: The equal segments and equal angles created by the equal sides AD and AE force the two outer triangles to match up exactly.

Exam Tip: Convert the angle equality \( \angle ADE = \angle AED \) into the angles you need for the congruence statement. Be explicit about which angle and side pair you are comparing.

Question 15(a). In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that △ADE ≅ △BCE and hence, AEB is an isosceles triangle.
Answer: From the figure, \( \angle ADE = \angle ADC + \angle CDE = 90° + 60° = 150° \) (since ABCD is a square with right angles, and CDE is equilateral with 60° angles). Similarly, \( \angle BCE = \angle BCD + \angle DCE = 90° + 60° = 150° \). Therefore, \( \angle ADE = \angle BCE \). In a square, opposite sides are equal, so \( AD = BC \). Since CDE is equilateral, \( DE = EC \). By SAS axiom, \( \triangle ADE \cong \triangle BCE \). From this congruence, the corresponding sides are equal: \( AE = BE \). Therefore, triangle AEB has two equal sides and is isosceles.
In simple words: The square's right angles plus the equilateral triangle's 60° angles create two larger triangles with matching angles and sides, which then form an isosceles triangle at the top.

Exam Tip: Calculate the composed angles (ADC + CDE and BCD + DCE) explicitly. Verify that all three congruence conditions (angle, side, side) are in place before concluding.

Question 15(b). In the figure (ii) given below, O is the point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that OCD is an isosceles triangle.
Answer: Since ABCD is a square, all sides are equal: \( AB = BC = CD = DA \). Since OAB is an equilateral triangle, \( OA = OB = AB \). Let the side length of the square be \( a \), so \( OA = OB = a \). In triangle OAD, we use the law of cosines or recognize that \( \angle OAD = \angle OAB + \angle BAD = 60° + 90° = 150° \). With \( OA = a \) and \( AD = a \), triangle OAD is isosceles with \( \angle OAD = 150° \). The base angles are equal: \( \angle AOD = \angle ADO = (180° - 150°)/2 = 15° \). By the same logic applied to triangle OBC: \( OB = a \), \( BC = a \), and \( \angle OBC = 60° + 90° = 150° \), so \( \angle BOC = \angle BCO = 15° \). Now, in triangle OCD, we have \( CD = a \) (side of square) and we can show that the angles at O and D work out such that \( OC = OD \) by using the symmetry of the construction or by direct calculation. Therefore, OCD is isosceles with \( OC = OD \).
In simple words: The equilateral triangle at the bottom forces point O to sit in a symmetric position relative to the top side, making it equidistant from the two top corners, which then creates an isosceles triangle at the top.

Exam Tip: Use the square's right angles and the equilateral triangle's 60° angles to build composed angles. Exploit symmetry: if the configuration is symmetric about a vertical line through the square's center, then OC = OD follows directly.

Question 15(c). In the adjoining figure, ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
Answer: In triangle ABC, ∠A = 90° and AB = AC. Since the two sides are equal, the angles opposite to them are also equal, so ∠B = ∠C. Using the angle sum property: 90° + ∠B + ∠B = 180°, which gives ∠B = 45°. Since AD bisects angle A, we have ∠BAD = ∠CAD = 45°. Now AD is common to both triangles ABD and ACD. We find ∠ACD = ∠ABD = 45°. By the AAS axiom, triangle ABD is congruent to triangle ACD. From this congruence, BD = CD. In triangle ABD, since ∠BAD = ∠ABD = 45°, the sides opposite these angles are equal, so AD = BD. Therefore, BC = BD + CD = BD + BD = 2BD. Substituting AD for BD, we get BC = 2AD.
In simple words: When you split angle A in half, it creates two matching triangles. The bottom side gets divided into two equal pieces, and each piece equals the height line AD. So the whole bottom side is twice the height.

Exam Tip: Use angle bisector properties and congruent triangles systematically - clearly state each angle equality and apply the AAS criterion explicitly.

Exercise 9.4

Question 1. In △PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is longest? Give reason for your answer.
Answer: The sum of all angles in a triangle is 180°. Therefore: ∠P + ∠Q + ∠R = 180°, which gives 70° + ∠Q + 30° = 180°, so ∠Q = 80°. The side opposite to the largest angle in any triangle is always the longest side. Since ∠Q = 80° is the greatest angle, side PR (which is opposite to ∠Q) is the longest side.
In simple words: The biggest angle is 80°. The side across from the biggest angle is always the longest, so PR is the longest side.

Exam Tip: Always find the third angle first using the angle sum property, then identify the largest angle and name the opposite side clearly.

Question 2. Show that in a right angled triangle, the hypotenuse is the longest side.
Answer: Consider a right triangle ABC where ∠B = 90°. The other two angles, ∠A and ∠C, must both be acute angles (less than 90°) because the sum of all three angles equals 180°. Since ∠B = 90° is larger than both ∠A and ∠C, it is the greatest angle in the triangle. The side opposite the greatest angle is always the greatest side. The side opposite to the 90° angle at B is AC, which is the hypotenuse. Therefore, AC is longer than both BC and AB, proving that the hypotenuse is the longest side in a right angled triangle.
In simple words: The right angle (90°) is always the biggest angle. The side opposite the biggest angle is always the longest, so the hypotenuse must be the longest side.

Exam Tip: Begin by establishing that the right angle is the largest angle, then apply the angle-side relationship to conclude about the hypotenuse.

Question 3. PQR is a right angle triangle at Q and PQ : QR = 3 : 2. Which is the least angle?
Answer: Since the triangle is right-angled at Q, we have ∠Q = 90°. The ratio PQ : QR = 3 : 2 tells us that PQ is the longer of the two legs. We can write PQ = 3x and QR = 2x for some positive value x. The smallest side in any triangle is opposite the smallest angle. Since QR = 2x is smaller than PQ = 3x, the angle opposite to QR is the smallest. The angle opposite side QR is ∠P. Therefore, ∠P is the least angle.
In simple words: The side QR is the shortest side because 2 is less than 3. The smallest angle is always opposite the shortest side, so ∠P is the smallest angle.

Exam Tip: Match sides to their opposite angles using the side-angle relationship - the smallest side is opposite the smallest angle.

Question 4. In △ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Which is
(i) the greatest angle?
(ii) the smallest angle?
Answer:
(i) Comparing the three sides: AB = 8 cm is the longest side. In any triangle, the angle opposite the longest side is the greatest angle. AB is opposite to angle C. Therefore, ∠C is the greatest angle.
(ii) Among the three sides, BC = 5.6 cm is the shortest. The angle opposite the shortest side is always the smallest angle. BC is opposite to angle A. Therefore, ∠A is the smallest angle.
In simple words: The longest side AB is opposite angle C, so C is the biggest angle. The shortest side BC is opposite angle A, so A is the smallest angle.

Exam Tip: Compare all three side lengths first, then use the side-angle relationship: longest side - greatest angle; shortest side - smallest angle.

Question 5. In a △ABC, ∠A = 50°, ∠B = 60°. Arrange the sides of the triangle in ascending order.
Answer: Using the angle sum property: ∠A + ∠B + ∠C = 180°, we get 50° + 60° + ∠C = 180°, so ∠C = 70°. Ordering the angles from smallest to largest: ∠A = 50° < ∠B = 60° < ∠C = 70°. In any triangle, when angles are in increasing order, the sides opposite to those angles are also in increasing order. The side opposite ∠A is BC, the side opposite ∠B is CA, and the side opposite ∠C is AB. Therefore, the sides in ascending order are BC < CA < AB.
In simple words: Find the third angle first: it's 70°. The angles from small to big are 50°, 60°, 70°. The sides opposite them go small to big in the same order: BC, CA, AB.

Exam Tip: Always compute the missing angle, order all angles, then name the opposite sides in the same order.

Question 6. In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show that
(i) BD > AD
(ii) DC > AD
(iii) AC > DC
(iv) AB > BD.
Answer: First, we find ∠A. Since ∠A + ∠B + ∠C = 180°, we have ∠A = 180° - 30° - 40° = 110°. Since AD bisects ∠A, we get ∠BAD = ∠CAD = 55°.
(i) In triangle ABD, ∠BAD = 55° and ∠B = 30°. Since ∠BAD > ∠B, the side opposite ∠BAD (which is BD) is greater than the side opposite ∠B (which is AD). Therefore, BD > AD.
(ii) In triangle ACD, ∠CAD = 55° and ∠C = 40°. Since ∠CAD > ∠C, the side opposite ∠CAD (which is DC) is greater than the side opposite ∠C (which is AD). Therefore, DC > AD.
(iii) In triangle ACD, the third angle is ∠ADC = 180° - 40° - 55° = 85°. Since ∠ADC = 85° > ∠CAD = 55°, the side opposite ∠ADC (which is AC) is greater than the side opposite ∠CAD (which is DC). Therefore, AC > DC.
(iv) In triangle ABD, ∠ADB = 180° - ∠ADC = 180° - 85° = 95°. Since ∠ADB = 95° > ∠BAD = 55°, the side opposite ∠ADB (which is AB) is greater than the side opposite ∠BAD (which is BD). Therefore, AB > BD.
In simple words: The angle bisector splits the 110° angle into two 55° pieces. Compare angles in each smaller triangle and use the rule: bigger angle, bigger opposite side.

Exam Tip: Find all angles in each sub-triangle formed by the bisector, then systematically compare angles within each triangle to deduce side relationships.

Question 7. In the adjoining figure, AD bisects ∠A. Arrange AB, BD and DC in the descending order of their lengths.
Answer: From the figure, ∠B = 60° and ∠C = 40°, so ∠A = 180° - 60° - 40° = 80°. Since AD bisects ∠A, each half is 40°. Using the exterior angle theorem on triangle ACD: ∠ADB (the exterior angle at D) equals the sum of the two non-adjacent interior angles, so ∠ADB = ∠DAC + ∠C = 40° + 40° = 80°. In triangle ABD, the angles are ∠BAD = 40°, ∠ABD = 60°, and ∠ADB = 80°. Ordering them: ∠BAD < ∠ABD < ∠ADB, which gives BD < AD < AB. In triangle ADC, ∠DAC = 40° = ∠C, which means triangle ADC is isosceles with AD = DC. Combining both results: BD < DC = AD < AB, so in descending order: AB > DC > BD.
In simple words: The angle bisector creates two triangles. In the first one, the angles tell us BD is smallest and AB is biggest. In the second one, the two equal angles mean AD equals DC. Putting it together: AB > DC > BD.

Exam Tip: Use the exterior angle theorem and isosceles triangle properties (equal angles mean equal opposite sides) to establish side comparisons across both sub-triangles.

Question 8(a). In the figure (1) given below, prove that (i) CF > AF (ii) DC > DF.
Answer:
(i) Using the angle sum property in triangle ABC with ∠A = 60° and ∠B = 65°, we get ∠ACB = 55°. In triangle BEC with ∠B = 65° and ∠BEC = 90°, we get ∠BCE = 25°. In triangle DFC with ∠D = 90° and ∠DCF = 25°, we get ∠DFC = 65°. The angles ∠AFE and ∠DFC are vertically opposite, so ∠AFE = 65°. By the exterior angle theorem applied to triangle AFC, ∠AFE = ∠FAC + ∠FCA. This gives 65° = ∠FAC + 30°, so ∠FAC = 35°. In triangle AFC, ∠FAC = 35° > ∠FCA = 30°, so the side opposite ∠FAC (which is CF) is greater than the side opposite ∠FCA (which is AF). Therefore, CF > AF.
(ii) In triangle DFC, ∠DFC = 65° > ∠DCF = 25°. The side opposite ∠DFC (which is DC) is therefore greater than the side opposite ∠DCF (which is DF). Thus, DC > DF.
In simple words: Find all angles by using the angle sum property in each smaller triangle. Compare angles within each triangle and use the angle-side relationship to show which sides are longer.

Exam Tip: Systematically find every angle in all triangles involved, use vertically opposite angles and the exterior angle theorem when needed, then apply the side-angle comparison rule.

Question 8(b). In the figure (2) given below, AB = AC. Prove that AB > CD.
Answer: Since AB = AC, the triangle ABC is isosceles. The angles opposite the equal sides are equal, so ∠ABC = ∠ACB = 70°. From the figure, points C and D are collinear with B, and ∠ACB and ∠ACD form a linear pair (they add to 180°). Therefore, ∠ACD = 180° - 70° = 110°. In triangle ACD, ∠CAD = 180° - 110° - 40° = 30° (since ∠ADC = 40° from the figure). Comparing triangle ABC and triangle ACD: in triangle ABC, ∠ABC = 70° is opposite side AC; in triangle ACD, ∠ADC = 40° is opposite side AC. Since ∠ABC > ∠ADC, and side AC is common to both, we can see that in triangle ABC, side AB is opposite ∠ACB = 70°, while in triangle ACD, side CD is opposite ∠CAD = 30°. Since 70° > 30°, and both sides are measured relative to points on the same line through C, AB > CD.
In simple words: Triangle ABC is isosceles with base angles of 70° each. When you extend the base to point D, the angle ∠ACD becomes 110°. In the new triangle ACD, all angles are smaller, so its sides are also smaller. AB is larger than CD.

Exam Tip: Use the isosceles triangle property first, then carefully construct angles on a straight line to compare angles in different triangles sharing a vertex.

Question 8(c). In the figure (3) given below, AC = CD. Prove that BC < CD.
Answer: Looking at the figure, angles ACB and ACD form a linear pair on line BD, so they sum to 180°. Since angle ACB is 70°, we get angle ACD = 110°. In triangle ACD, because AC equals CD, the angles opposite these equal sides must also be equal. Setting angle CAD = angle CDA = x, and using the angle sum property of triangles (x + x + 110° = 180°), we find x = 35°. Therefore angle CAD = 35°. From the figure, angle CAD and angle CAB together form the 70° angle at vertex A, so angle CAB = 70° - 35° = 35°. In triangle ABC, the angle sum gives us angle ABC = 180° - 70° - 35° = 75°. Since angle ABC (75°) is greater than angle CAB (35°), the side AC opposite the larger angle must be longer than side BC opposite the smaller angle. Therefore AC > BC. Because we are given that AC = CD, we can conclude CD > BC, which means BC < CD.
In simple words: When AC equals CD, the angles opposite these sides are also equal. Using this fact and comparing angles in triangle ABC, we can show that side BC must be shorter than side CD.

Exam Tip: Always remember that when two sides of a triangle are equal, their opposite angles are equal. Use this property along with the angle sum rule to establish relationships between different line segments.

Question 9(a). In the figure (i) given below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
Answer: Consider triangles AOB and COD, where O is the intersection point. In triangle AOB, since angle B is smaller than angle A, the side opposite to the smaller angle must be smaller. Therefore, AO is smaller than BO. This gives us our first relationship: AO < BO. Similarly, in triangle COD, because angle C is smaller than angle D, the side opposite angle C (which is OD) must be smaller than the side opposite angle D (which is OC). So OD < OC. Adding these two inequalities together: AO + OD < BO + OC. The left side simplifies to AD and the right side simplifies to BC. Therefore, AD < BC.
In simple words: When an angle in a triangle is smaller, the side across from it is also smaller. By combining this rule for two different triangles and adding the results, we can prove that AD must be shorter than BC.

Exam Tip: Break the problem into separate triangles and apply the angle-side relationship in each one. Then combine your findings using addition or subtraction to reach the final result.

Question 9(b). In the figure (ii) given below, D is any point on the side BC of △ABC. If AB > AC, show that AB > AD.
Answer: We are given that AB is longer than AC. Since the angle opposite a longer side is larger, angle ACB must be greater than angle ABC. Next, we use the exterior angle theorem. Angle ADB is an exterior angle to triangle ACD, so it equals the sum of the two non-adjacent interior angles: angle ADB = angle ACD + angle DAC. This means angle ADB is definitely larger than angle ACD (since we are adding a positive angle to it). We established that angle ACD > angle ABC, so angle ADB > angle ABC. In triangle ABD, since angle ADB is greater than angle ABD, the side opposite the larger angle must be longer. Therefore, AB > AD.
In simple words: An exterior angle of a triangle is always bigger than any non-adjacent interior angle. Using this fact, we can show that a certain angle in triangle ABD is large enough to guarantee that side AB is longer than side AD.

Exam Tip: The exterior angle theorem is a powerful tool - remember that an exterior angle equals the sum of the two remote interior angles, making it larger than either one individually.

Question 10(i). Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer.
Answer: For any triangle, the sum of any two sides must be strictly greater than the third side. Let us check all three combinations. Adding the two shorter sides: 4 cm + 3 cm = 7 cm. This sum equals the third side rather than being greater than it. Since the triangle inequality condition fails (we need the sum to be strictly greater, not equal), a triangle with these side lengths cannot be formed.
In simple words: A triangle can only exist if any two sides added together make a length longer than the remaining side. Here, 4 + 3 only equals 7, so no triangle is possible.

Exam Tip: Always test all three possible combinations of adding two sides and comparing to the third side. Even if two combinations pass, one failure means the triangle cannot exist.

Question 10(ii). Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.
Answer: For a valid triangle, the sum of any two sides must be strictly greater than the remaining side. Let us verify this condition. Taking the two shorter sides: 9 cm + 7 cm = 16 cm. However, the third side is 17 cm, which is larger than 16 cm. Since the sum of the two shorter sides (16 cm) is less than the third side (17 cm), the triangle inequality is violated. Therefore, a triangle with sides 9 cm, 7 cm, and 17 cm cannot be constructed.
In simple words: When two shorter sides add up to less than the longest side, you cannot build a triangle. Here 9 + 7 = 16, which is smaller than 17, so this triangle is impossible.

Exam Tip: Focus on checking whether the sum of the two smaller sides exceeds the largest side—this is often the key test that determines whether a triangle can be formed.

Question 10(iii). Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer.
Answer: A triangle can be constructed only if the sum of any two sides is greater than the remaining side. We check all three conditions: Adding the smallest two sides, 7 cm + 4 cm = 11 cm, which exceeds 8 cm ✓. Adding 8 cm and 4 cm gives 12 cm, which exceeds 7 cm ✓. Adding the two largest sides, 8 cm + 7 cm = 15 cm, which exceeds 4 cm ✓. Since all three triangle inequality conditions are satisfied, a triangle with sides 8 cm, 7 cm, and 4 cm can definitely be constructed.
In simple words: This set of side lengths passes the triangle test - when you add any two sides together, the result is always bigger than the third side. So a real triangle can be made.

Exam Tip: When all three combinations satisfy the triangle inequality (sum of two sides > third side), construction is possible. Write out each check clearly to show your reasoning completely.

Question 1. Which of the following is not a criterion for congruency of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Answer: (c) SSA
In simple words: Three valid rules exist for proving triangles are congruent: SAS (two sides and the angle between them), ASA (two angles and the side between them), and SSS (all three sides). SSA (two sides and a non-included angle) does not guarantee congruence and is therefore not a valid criterion.

Exam Tip: Memorise the four main congruence criteria (SAS, ASA, SSS, AAS) and understand why SSA fails - the angle not between the two sides leaves ambiguity about triangle shape.

Question 2. In the adjoining figure, AB = FC, EF = BD and ∠AFE = ∠CBD. Then the rule by which △AFE ≅ △CBD is
(a) SAS
(b) ASA
(c) SSS
(d) AAS
Answer: (a) SAS
In simple words: We are given two sides and the angle between them in both triangles. The segment AB equals CF, and adding the same length BF to both sides shows AF equals BC. We also know EF equals BD. The angles AFE and CBD are equal. This gives us two sides and the included angle, which is the SAS rule.

Exam Tip: When two equal sides and an equal included angle are identified between two triangles, always choose SAS as the congruence criterion. The word "included" is crucial—the angle must be between the two equal sides.

Question 3. In the adjoining figure, AB ⊥ BE and FE ⊥ BE. If AB = FE and BC = DE, then
(a) △ABD ≅ △EFC
(b) △ABD ≅ △FEC
(c) △ABD ≅ △ECF
(d) △ABD ≅ △CEF
Answer: (b) △ABD ≅ △FEC
In simple words: Since AB is perpendicular to BE and FE is perpendicular to BE, both angles ABD and FEC are right angles (90°). We are told BC equals DE. Adding CD to both sides gives us BD equals EC. Combined with AB equals FE and the right angles at B and E, the SAS condition is satisfied: two sides (AB = FE and BD = EC) with an included right angle.

Exam Tip: Perpendicular marks in a figure guarantee right angles. Use these right angles as one part of your SAS or other congruence criteria. Pay careful attention to which vertices correspond when stating triangle congruence.

Question 4. In the adjoining figure, AB = AC and AD is median of △ABC, then ∠ADC is equal to
(a) 60°
(b) 120°
(c) 90°
(d) 75°
Answer: (c) 90°
In simple words: Since AB equals AC, triangle ABC is isosceles. A median from A goes to D, the midpoint of BC, so BD equals DC. Now we have: AB = AC, BD = DC (since D is the midpoint), and AD = AD (common side). By the SSS congruence rule, triangles ADB and ADC are congruent. Since these triangles are congruent, their corresponding angles must be equal, so angle ADB = angle ADC. These two angles form a straight line along BC, so they sum to 180°. If angle ADB = angle ADC and angle ADB + angle ADC = 180°, then each angle equals 90°.

Exam Tip: In an isosceles triangle, the median from the vertex angle is also the perpendicular bisector of the base. This is a useful property to remember—it directly tells us that ∠ADC = 90°.

Question 5. In the adjoining figure, O is the mid-point of AB. If ∠ACO = ∠BDO, then ∠OAC is equal to
(a) ∠OCA
(b) ∠ODB
(c) ∠OBD
(d) ∠BOD
Answer: (c) ∠OBD
In simple words: Since O is the midpoint of AB, we have AO = OB. We are given that angle ACO equals angle BDO. Additionally, angles AOC and BOD are vertically opposite angles (formed where two lines cross), so they are equal. With AO = OB, one pair of equal angles, and another pair of vertically opposite angles, triangles AOC and BOD satisfy the AAS (Angle-Angle-Side) congruence criterion. Therefore, all corresponding parts of these congruent triangles are equal, which means angle OAC = angle OBD.

Exam Tip: Vertically opposite angles are always equal—use this fact freely in congruence proofs. When you have equal sides and equal angles positioned as described, identify which congruence criterion applies and match corresponding parts carefully.

Question 6. In the adjoining figure, AC = BD. If ∠CAB = ∠DBA, then ∠ACB is equal to
(a) ∠BAD
(b) ∠ABC
(c) ∠ABD
(d) ∠BDA
Answer: (d) ∠BDA
In simple words: In triangles CAB and DBA, we know AC = DB (given), AB = AB (it is the common side to both triangles), and angle CAB = angle DBA (given). These three pieces of information—two sides and the included angle—match the SAS congruence criterion. Therefore, triangle CAB is congruent to triangle DBA. Since the triangles are congruent, their corresponding angles must be equal. The angle ACB in triangle CAB corresponds to angle BDA in triangle DBA, so angle ACB = angle BDA.

Exam Tip: Be careful about the order of vertices when stating congruence—△CAB ≅ △DBA tells you which angles match. Always identify the correspondence correctly to match the right pairs of angles.

Question 7. In the adjoining figure, ABCD is a quadrilateral in which BN and DM are drawn perpendiculars to AC such that BN = DM. If OB = 4 cm, then BD is
(a) 4 cm
(b) 8 cm
(c) 12 cm
(d) 16 cm

Question 8. In △ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Answer: (b) 50°
In simple words: When two sides of a triangle are equal, the angles opposite those sides are also equal. Since AB = AC, the angles at B and C must be the same.

Exam Tip: In an isosceles triangle, always identify which two sides are equal, then remember that the angles facing those equal sides will match.

Question 9. In △ABC, BC = AB and ∠B = 80°. Then ∠A is equal to
(a) 80°
(b) 40°
(c) 50°
(d) 100°
Answer: (c) 50°
In simple words: Since BC = AB, the angles at A and C are equal. Let both equal x. Using the angle sum rule, x + 80° + x = 180°, so 2x = 100°, giving x = 50°.

Exam Tip: Always use the fact that the sum of all angles in any triangle equals 180° to find missing angles.

Question 10. In △PQR, ∠R = ∠P, QR = 4 cm and PR = 5 cm. Then the length of PQ is
(a) 4 cm
(b) 5 cm
(c) 2 cm
(d) 2.5 cm
Answer: (a) 4 cm
In simple words: When two angles in a triangle are equal, the sides facing those angles must also be equal. Since ∠R = ∠P, the side opposite ∠R (which is PQ) equals the side opposite ∠P (which is QR). So PQ = 4 cm.

Exam Tip: Remember - equal angles are always matched with equal opposite sides in a triangle.

Question 11. In △ABC and △PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but isosceles
(d) neither congruent nor isosceles
Answer: (a) isosceles but not congruent
In simple words: Triangle ABC is isosceles because AB = AC. Since ∠B = ∠C in an isosceles triangle, and we're told ∠B = ∠Q and ∠C = ∠P, triangle PQR must also be isosceles with PR = QR. However, the two triangles do not have to be congruent just from this information.

Exam Tip: Don't confuse having equal angles with being congruent - congruence requires matching shape AND size, while equal angles alone only guarantee shape.

Question 12. Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle can not be
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm
Answer: (d) 3.4 cm
In simple words: For any triangle, the third side must be larger than the difference of the other two sides. Here, 5 - 1.5 = 3.5 cm. Since 3.4 cm is smaller than 3.5 cm, a triangle with sides 5 cm, 1.5 cm, and 3.4 cm cannot exist.

Exam Tip: Use the triangle inequality rule: the third side must always be greater than (larger side - smaller side) and less than (larger side + smaller side).

Question 13. If a, b, c are lengths of the sides of a triangle then,
(a) a - b > c
(b) c > a + b
(c) c = a + b
(d) c < a + b
Answer: (d) c < a + b
In simple words: This is the triangle inequality rule. In any triangle, one side must always be smaller than the sum of the other two sides.

Exam Tip: The triangle inequality has three forms: a + b > c, b + c > a, and a + c > b - all must be true for a valid triangle.

Question 14. It is not possible to construct a triangle when the lengths of its sides are
(a) 6 cm, 7 cm, 8 cm
(b) 4 cm, 6 cm, 6 cm
(c) 5.3 cm, 2.2 cm, 3.1 cm
(d) 9.3 cm, 5.2 cm, 7.4 cm
Answer: (c) 5.3 cm, 2.2 cm, 3.1 cm
In simple words: For these sides, 2.2 + 3.1 = 5.3. Since the sum of two sides equals the third side rather than being greater, this violates the triangle inequality rule.

Exam Tip: Check all three triangle inequality conditions. If the sum of any two sides equals or is less than the third side, construction is impossible.

Question 15. In △PQR, if ∠R > ∠Q, then
(a) QR > PR
(b) PQ > PR
(c) PQ < PR
(d) QR < PR
Answer: (b) PQ > PR
In simple words: In any triangle, the side facing a larger angle is always longer. Since ∠R is larger than ∠Q, the side opposite to ∠R (which is PQ) must be longer than the side opposite to ∠Q (which is PR).

Exam Tip: Larger angles always face longer sides, and smaller angles always face shorter sides - this is a fundamental triangle property.

Question 16. If triangle PQR is right angled at Q, then
(a) PR = PQ
(b) PR < PQ
(c) PR < QR
(d) PR > PQ
Answer: (d) PR > PQ
In simple words: In a right-angled triangle, the side opposite the right angle (called the hypotenuse) is always the longest side. Here, PR is the hypotenuse, so it must be larger than PQ.

Exam Tip: The hypotenuse in a right triangle is always longer than either of the other two sides (the perpendicular and the base).

Question 17. If triangle ABC is obtuse angled and ∠C is obtuse, then
(a) AB > BC
(b) AB = BC
(c) AB < BC
(d) AC > AB
Answer: (a) AB > BC
In simple words: When ∠C is obtuse (the largest angle in the triangle), the side facing it, which is AB, must be the longest side. So AB is greater than BC.

Exam Tip: The obtuse angle in a triangle is always the largest, and it always faces the longest side.

Question 18. If the lengths of two sides of an isosceles triangle are 4 cm and 10 cm, then the length of the third side is
(a) 4 cm
(b) 10 cm
(c) 7 cm
(d) 14 cm
Answer: (b) 10 cm
In simple words: In an isosceles triangle, two sides must be equal. The third side could be either 4 cm or 10 cm. If it's 4 cm, then 4 + 4 = 8, which is less than 10 - this breaks the triangle rule. So the third side must be 10 cm.

Exam Tip: For isosceles triangles, always check both possibilities against the triangle inequality before deciding which is valid.

Question 19. Consider the following two statements:
Statement 1: If two angles and one side of a triangle is equal to two angles and one side of another triangle then the two triangles are congruent.
Statement 2: The angle opposite to equal sides of an isosceles triangle are equal.
Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (a) Both the statements are true.
In simple words: Statement 1 describes the AAS (Angle-Angle-Side) and ASA (Angle-Side-Angle) congruence rules, which are both valid. Statement 2 is also a fundamental property of isosceles triangles - the base angles are always equal.

Exam Tip: Know all five congruence rules (SSS, SAS, ASA, AAS, RHS) and the key property that base angles of an isosceles triangle are always equal.

Question. Assertion (A): If two triangles are congruent to each other then the ratio of corresponding side is 1 : 1.
Reason (R): Two triangles are congruent if and only if they have same shape and size.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: The reason gives the definition of congruent triangles. When two triangles are truly congruent (same shape and size), all their matching sides are equal in length, so their ratio is 1:1. The reason directly explains why the assertion is true.

Exam Tip: In assertion-reason questions, check both statements first, then verify whether the reason logically supports the assertion.

Question. Assertion (A): In the adjoining figure, △ABC ≅ △PQR.
Reason (R): Two triangles are congruent if two angles and included side of one triangle are equal to two angles and the included side of other triangle.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: From the figure, ∠B = ∠Q = 45°, BC = QR = 4 cm, and ∠C = ∠R = 50°. This matches the ASA (Angle-Side-Angle) rule stated in the reason, so both triangles are definitely congruent.

Exam Tip: When using ASA or AAS, always verify that the side is truly included (between the two angles) before applying the rule.

Question. Assertion (A): Two triangles are congruent to each other by RHS rule of congruency. The base of the one triangle is 6 cm and the height is 4 cm. The sum of area of two triangle is 24 cm².
Reason (R): Two triangles are congruent if they have same shape and size. Congruent triangles have equal area.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: When two right triangles are congruent by RHS, they have identical shape and size, making their areas equal. Area of one triangle = ½ × 6 × 4 = 12 cm². Sum of both = 12 + 12 = 24 cm². The reason correctly explains why this sum is what it is.

Exam Tip: Congruent shapes always have equal areas. If told triangles are congruent, you can immediately conclude they have the same area.

Question. Congruent figures are identical in every respect, meaning they have the same shape and occupy the same amount of space. If two triangles have the same shape and size (i.e., they are congruent), then they must necessarily cover the same amount of space, meaning they have equal areas. What does this tell us about the relationship between congruent triangles and their areas?
Answer: When two triangles match perfectly in every way - same shape and same measurements - they also cover the same amount of surface. This is because congruence means the triangles are exactly alike, so they take up equal space. The area formula for a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). For example, if one triangle has a base of 6 cm and a height of 4 cm, its area is \( \frac{1}{2} \times 6 \times 4 = 12 \text{ cm}^2 \). When two triangles are congruent, their areas must be the same, so the second triangle also has an area of 12 cm². If you add both areas together, you get 12 cm² + 12 cm² = 24 cm².
In simple words: Matching triangles cover the same amount of space and have the same area.

Exam Tip: Remember that congruent shapes always have equal areas - this is a key property that often appears in exam questions about comparing two identical figures.

Question 4. Assertion (A): If △ABC ≅ △PQR and AB = 4 units, then AB² + PQ² = 32. Reason (R): When two triangles are congruent to each other, then their corresponding sides are equal in length.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: When two triangles are perfectly identical, sides that match up with each other are always the same length. This basic rule is called CPCTC (Corresponding Parts of Congruent Triangles are Congruent). Since triangle ABC matches triangle PQR, the side AB in the first triangle matches PQ in the second triangle. Because both equal 4 units, we can work out: 4² + 4² = 16 + 16 = 32, which proves the assertion is true and the reason explains it correctly.

Exam Tip: Assertion-Reason questions test whether you understand not just facts, but also why those facts are true - always verify both the statement AND its explanation.

Question 5. Assertion (A): △ABC ≅ △PQR. If 2AB = 3BC = 4CA = 12 cm, then the sum of perimeter of both triangle is 26 cm. Reason (R): If two triangles are congruent to each other, then their corresponding sides are equal in length.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: First, find each side length from the given information. From 2AB = 12, we get AB = 6 cm. From 3BC = 12, we get BC = 4 cm. From 4CA = 12, we get CA = 3 cm. Add them up: the perimeter of triangle ABC is 6 + 4 + 3 = 13 cm. Because the two triangles are congruent, triangle PQR has the same perimeter - also 13 cm. Together: 13 + 13 = 26 cm. The reason is correct because congruent triangles always have matching corresponding sides and equal perimeters.

Exam Tip: When given ratios like "2AB = 3BC = 4CA", always solve for each individual side first before finding the perimeter or sum.

Chapter Test

Question 1. In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.
Answer: The two triangles will not be congruent. The reason is that AB and EF are not matching sides. Even though the angles at A and D are the same, and the angles at B and E are the same, the side AB in triangle ABC does not match with side EF in triangle DEF in terms of which angles they join. For triangles to be congruent using the ASA (Angle-Side-Angle) rule, the shared side must connect the two matching angles in both triangles. Here, AB connects angles A and B in the first triangle, but EF connects angles E and F (not D and E) in the second triangle. Because the corresponding parts do not match up correctly, we cannot say the triangles are congruent.
In simple words: Two triangles are only congruent when matching angles and matching sides are in the same positions. Here they are not, so these triangles cannot be proven congruent.

Exam Tip: Always identify which sides connect which angles - correct matching of angle-side pairs is essential for proving triangle congruence.

Question 2. In the adjoining figure, ABCD is a square. P, Q and R are points on the sides AB, BC and CD respectively such that AP = BQ = CR and ∠PQR = 90°. Prove that (a) △PBQ ≅ △QCR (b) PQ = QR (c) ∠PRQ = 45°
Answer: (a) In triangles PBQ and QCR: - Side BQ = CR (given that BQ = CR) - Since ABCD is a square: AB = BC. Also, AP = BQ (given). So: AB - AP = BC - BQ, which means PB = QC. - Angle PBQ = Angle QCR (both equal 90° because each angle in a square is 90°) - Therefore, △PBQ ≅ △QCR by the SAS (Side-Angle-Side) axiom. (b) Since the two triangles are congruent (proved above), their matching sides must be equal. The side PQ in triangle PBQ corresponds to side QR in triangle QCR. Therefore, PQ = QR. (c) In triangle PQR, we know PQ = QR (proved in part b) and ∠Q = 90° (given). This means triangle PQR is isosceles with two equal sides and a right angle between them. Let ∠P = ∠R = x. The sum of all angles in a triangle equals 180°, so: x + 90° + x = 180°, which gives 2x = 90°, therefore x = 45°. This proves ∠PRQ = 45°.
In simple words: When you have two triangles with matching sides and a matching angle between those sides, they are congruent. Congruent triangles have all matching sides equal. In an isosceles right triangle where the right angle is in the middle, the two base angles must each be 45°.

Exam Tip: For multi-part proofs, work through each section in order - earlier parts often provide the facts you need for later parts. Always state which congruence rule (SAS, ASA, SSS) you are using and why it applies.

Question 3. In the adjoining figure, OA ⊥ OD, OC ⊥ OB, OD = OA and OB = OC. Prove that AB = CD.
Answer:Looking at the figure, we know: - ∠AOD = 90° (since OA ⊥ OD) - ∠COB = 90° (since OC ⊥ OB) Adding ∠AOC to both sides of the equation ∠AOD = ∠COB: - ∠AOD + ∠AOC = ∠COB + ∠AOC - This simplifies to: ∠COD = ∠AOB Now, compare triangles AOB and DOC: - OA = OD (given) - OB = OC (given) - ∠AOB = ∠COD (proved above) By the SAS (Side-Angle-Side) axiom, △AOB ≅ △DOC. Since corresponding parts of congruent triangles are equal, AB = CD.
In simple words: When two angles are equal and you add the same angle to both of them, the results stay equal. By showing that matching angles and matching sides are the same in both triangles, you can prove the triangles are congruent, which means their remaining sides must also be equal.

Exam Tip: When dealing with perpendicular lines, remember that each creates a 90° angle. Adding or subtracting angles strategically can help you find equal angles needed for congruence proofs.

Question 4. In the adjoining figure, PQ || BA and RS || CA. If BP = RC, prove that: (i) △BSR ≅ △PQC (ii) BS = PQ (iii) RS = CQ.
Answer:Given that BP = RC: - Subtract PR from both sides: BP - PR = RC - PR - This gives us: BR = PC Now compare triangles BSR and PQC: - ∠B = ∠P (these are corresponding angles because PQ || BA) - ∠R = ∠C (these are corresponding angles because RS || CA) - BR = PC (proved above) By the ASA (Angle-Side-Angle) axiom, △BSR ≅ △PQC. (ii) Since the two triangles are congruent, their matching sides must be equal. The side BS in triangle BSR corresponds to side PQ in triangle PQC. Therefore, BS = PQ. (iii) Similarly, the side RS in triangle BSR corresponds to side CQ in triangle PQC. Therefore, RS = CQ.
In simple words: Parallel lines create equal angles when cut by the same line. By finding matching angles and a matching side between two triangles, you can prove they are congruent. Once triangles are congruent, all their matching parts are automatically equal.

Exam Tip: Properties of parallel lines (corresponding angles, alternate angles) are key tools for finding equal angles in congruence proofs - always identify all parallel lines given in the problem first.

Question 5. In the adjoining figure, AB = AC, D is a point in the interior of △ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of △ABC.
Answer:Given: AB = AC and ∠DBC = ∠DCB Step 1: Since AB = AC, triangle ABC is isosceles. In an isosceles triangle, angles opposite to equal sides are equal. Therefore, ∠ABC = ∠ACB. Step 2: We are given that ∠DBC = ∠DCB. In any triangle, when two angles are equal, the sides opposite to them are also equal. Therefore, DB = DC. Step 3: Now compare triangles ABD and ACD: - AB = AC (given) - DB = DC (proved in Step 2) - AD = AD (common side) By the SSS (Side-Side-Side) axiom, △ABD ≅ △ACD. Step 4: Since corresponding parts of congruent triangles are equal, ∠BAD = ∠CAD. This means the line AD divides angle BAC into two equal parts, so AD bisects ∠BAC.
In simple words: An isosceles triangle has two equal sides and two equal angles opposite those sides. When a point inside such a triangle creates equal angles at the base, the line from the top vertex to that point must divide the top angle equally into two parts.

Exam Tip: In isosceles triangle problems, always use the property that equal sides produce equal opposite angles - this often gives you the extra facts needed to prove congruence of smaller triangles formed by additional points.

Question 6. In the adjoining figure, AB || DC. CE and DE bisects ∠BCD and ∠ADC respectively. Prove that AB = AD + BC.
Answer:Since DE bisects ∠D (angle ADC): - ∠EDA = ∠EDC = ∠D/2 Since AB || CD and DE cuts these parallel lines, alternate interior angles are equal: - ∠DEA = ∠EDC = ∠D/2 In triangle ADE: - ∠ADE = ∠DEA = ∠D/2 When two angles of a triangle are equal, the sides opposite to them are also equal. Therefore, AD = AE. Similarly, since CE bisects ∠C (angle BCD): - ∠ECB = ∠ECD = ∠C/2 Since AB || CD and CE cuts these parallel lines: - ∠CEB = ∠ECD = ∠C/2 In triangle BCE: - ∠BEC = ∠BCE = ∠C/2 Therefore, BE = BC. Since AB = AE + EB and we have proved that AD = AE and BC = BE: - AB = AD + BC
In simple words: When a line bisects an angle, it splits the angle into two equal parts. Parallel lines cut by the same line create equal matching angles. Using these properties, you can show that certain triangles are isosceles (two equal sides), and then combine these equal sides to reach the final proof.

Exam Tip: Look for angle bisectors in parallel line problems - they often create isosceles triangles with one angle being half an original angle, which becomes the key to solving the proof.

Question 7. In △ABC, D is a point on BC such that AD is the bisector of ∠BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that △CAE is isosceles.
Answer: From the given information: - AD bisects ∠BAC, so ∠BAD = ∠CAD - CE is parallel to AD Since CE || AD, and BC (extended to E) is a transversal cutting these parallel lines: - ∠CAD and ∠ACE are alternate interior angles, so ∠ACE = ∠CAD Also, since CE || AD, and line AC is a transversal: - ∠DAC and ∠CEA are alternate interior angles, so ∠CEA = ∠DAC (which is the same as ∠CAD) Since ∠BAD = ∠CAD (AD bisects angle BAC), we have: - ∠ACE = ∠CAD = ∠CEA In triangle ACE: - ∠ACE = ∠CEA When two angles of a triangle are equal, the sides opposite to those angles are also equal. The side AE is opposite to ∠ACE, and side AC is opposite to ∠CEA. Therefore, AE = AC, which means triangle CAE is isosceles.
In simple words: When you have a parallel line and use angle bisector properties, you create two equal angles inside a triangle. A triangle with two equal angles always has two equal sides facing those angles, making it isosceles.

Exam Tip: Angle bisectors combined with parallel lines are a powerful combination for creating equal angles - always check for these patterns and use alternate interior angle properties to find matching angles.

Question 8. In the adjoining figure, ABC is a right angled triangle at B. ADEC and BCFG are squares. Prove that AF = BE.
Answer: Both ADEC and BCFG are squares. When we look at triangles BCE and FCA, we find that angle BCE equals angle ACF because each is made by adding 90 degrees to angle BCA. Also, AC and CE are sides of squares so they must be equal, and the same is true for BC and CF. Using the SAS rule, triangles BCE and FCA match exactly. Since matching triangles have matching sides, we get AF = BE.
In simple words: The two triangles formed by the squares are identical, so their matching sides must also be identical.

Exam Tip: Recognize that angles BCE and ACF are both 90 degrees added to the same base angle - this equality is key to the congruence proof.

Question 9. In the adjoining figure, TR = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that RB = SA.
Answer: Since angles 1 and 4 are vertically opposite, they must be equal. From the given relationships ∠1 = 2∠2 and ∠4 = 2∠3, we get ∠2 = ∠3. Since TR = TS, the angles opposite these equal sides must match: ∠TRS = ∠TSR. Subtracting ∠2 from the first and ∠3 from the second gives ∠ARB = ∠BSA. Angle RTB appears in both triangles RBT and SAT as a shared angle. By the ASA rule, triangles RBT and SAT are congruent. Matching sides of congruent triangles are equal, so RB = SA.
In simple words: The two triangles match using two angles and the side between them, so their other matching sides must be equal.

Exam Tip: The key step is showing ∠2 = ∠3 from the given angle relationships - this unlocks the rest of the proof.

Question 10(a). In the figure (1) given below, find the value of x.
Answer: The bottom angle at B is marked as 36 degrees, and the two tick marks on the base show BD = AB. When two sides of a triangle are equal, the angles opposite them must also be equal. So angle ABD = angle BAD = 36 degrees. Using the fact that all angles in a triangle add to 180 degrees, we get angle BDA = 180 - (36 + 36) = 108 degrees. The angles BDA and ADC form a straight line, so they add to 180 degrees: 108 + angle ADC = 180, giving angle ADC = 72 degrees. In triangle ADC, we have AD = AC (shown by the tick marks), so angle ADC = angle ACD = 72 degrees. Now, 72 + 72 + angle DAC = 180, so angle DAC = 36 degrees. Finally, angle BAD + angle DAC + x = 180, which gives 36 + 36 + x = 180, so x = 108 degrees.
In simple words: Mark equal sides carefully and use the equal-angles rule. When you find all the angles in the smaller triangle, add them up with the exterior angle to get your answer.

Exam Tip: Draw and label all angle measures as you work - this helps track which angles are equal and prevents arithmetic mistakes.

Question 10(b). In the figure (2) given below, AB = AC and DE || BC. Calculate (i) x (ii) y (iii) ∠BAC
Answer: (i) Since AB = AC, angles ABC and ACB must be equal. This means 2x = (y - 2). Rearranging gives y = 2x + 2. Because DE is parallel to BC, angle ADE and angle ABC are matching angles on the same side of a transversal, so they are equal: x + y - 36 = 2x, which simplifies to y - 36 = x. Plugging y = 2x + 2 into this equation: 2x + 2 - 36 = x, so x = 34.
(ii) Using y = 2x + 2 and substituting x = 34: y = 2(34) + 2 = 70.
(iii) The sum of angles in triangle ABC is 180 degrees. We have angle BAC + angle ABC + angle ACB = 180, which becomes angle BAC + 2x + (y - 2) = 180. Substituting our values: angle BAC + 68 + 68 = 180, so angle BAC = 44 degrees.
In simple words: Start with the equal sides to link two variables. Use parallel lines to find another connection. Then solve both equations together to get your unknowns.

Exam Tip: When a line is parallel to a side, corresponding angles are equal - use this rule to set up your second equation.

Question 10(c). In the figure (3) given below, calculate the size of each lettered angle.
Answer: Angles AEB and DEC are vertically opposite, so both equal 80 degrees. In triangle AEB, the angles must sum to 180: x + 54 + 80 = 180, giving x = 46 degrees. The sides AB and BC are equal (marked with ticks), so the angles opposite them must be equal: angle BAC = angle ACB = 54 degrees. In triangle ABC, the angle sum gives (x + y) + 54 + 54 = 180. Substituting x = 46: 46 + y + 108 = 180, so y = 26 degrees. For angle z, note that angles ECD and BAE are alternate angles on a line cutting two parallel lines, so angle ECD = 54 degrees. The angles on a straight line sum to 180: 54 + 54 + z = 180, giving z = 72 degrees.
In simple words: Vertically opposite angles are always equal. Equal sides mean equal opposite angles. Line angles always sum to 180.

Exam Tip: Identify all equal side pairs first, then use the rule that opposite angles in such triangles are equal to unlock the rest.

Question 11(a). In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Show that (i) AC > DC (ii) AB > AD.
Answer: (i) In triangle ADC, sides AD and DC are equal, so angles opposite them must be equal. Since angle ACD = 35 degrees, angle DAC must also equal 35 degrees. The remaining angle is angle ADC = 180 - (35 + 35) = 110 degrees. Angle ADC is larger than angle DAC, and the side opposite a larger angle is always longer. Therefore, AC (opposite the 110-degree angle) is greater than DC (opposite the 35-degree angle).
(ii) The angle ADB lies on the opposite side of DC from angle ADC, so angle ADB = 180 - 110 = 70 degrees. In triangle ABD, sides AD and BD are equal, making the angles opposite them equal. If we call each of these angles a, then a + a + 70 = 180, giving a = 55 degrees. Now angle ADB (70 degrees) is greater than angle ABD (55 degrees), so the side opposite it must be longer. This means AB > AD.
In simple words: When two sides are equal, their opposite angles are equal. A larger angle always sits opposite a longer side.

Exam Tip: Use the "larger angle, longer side" principle - once you identify which angle is bigger, you immediately know which side is longer.

Question 11(b). In the figure (2) given below, prove that (i) x + y = 90° (ii) z = 90° (iii) AB = BC.
Answer: (i) The line through A is parallel to line BC, so angle ACB equals angle CAE by the alternate angles rule. Let this angle be x. In triangle ABC, the angle sum is x + (y + y) + x = 180, which gives 2x + 2y = 180, so x + y = 90 degrees.
(ii) In triangle BCD, we have y + z + angle BCD = 180. Since angle BCD = x (from the parallel line rule), we get y + z + x = 180. From part (i), we know x + y = 90, so substituting: 90 + z = 180, giving z = 90 degrees.
(iii) In triangle ABC, angle ACB = angle BAC = x. Since these two angles of the triangle are equal, the sides opposite them must be equal, so AB = BC.
In simple words: Parallel lines create equal alternate angles. Equal angles in a triangle mean equal opposite sides.

Exam Tip: Parallel lines automatically give you angle equalities - look for these relationships first, then build your equations from there.

Question 12. In the adjoining figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that (i) △ABD ≅ △ACD (ii) △ABP ≅ △ACP (iii) AP bisects ∠A as well as ∠D (iv) AP is the perpendicular bisector of BC.
Answer: (i) Since ABC is isosceles with base BC, we have AB = AC. Since DBC is isosceles with base BC, we have DB = DC. The side AD is common to both triangles ABD and ACD. By the SSS rule, triangle ABD is congruent to triangle ACD.
(ii) From the congruence in part (i), the matching angles are equal: angle BAD = angle CAD and angle ADB = angle ADC. Also, angle ABD = angle ACD. In triangles ABP and ACP, we have AB = AC (from the isosceles triangle ABC), angle BAP = angle CAP (since AP bisects angle A), and AP is common. By the SAS rule, triangle ABP is congruent to triangle ACP.
(iii) From the congruence in part (i), angle BAD = angle CAD, which means AP bisects angle A. Similarly, angle BDP = angle CDP, so AP bisects angle D as well.
(iv) From the congruence in part (ii), we get BP = CP and angle APB = angle APC. Since these angles form a straight line, each must be 90 degrees. Therefore, AP is perpendicular to BC and passes through its midpoint, making AP the perpendicular bisector of BC.
In simple words: Two matching isosceles triangles on the same base create a line from their apex that cuts the base in half at a right angle.

Exam Tip: Label matching parts carefully - once you prove the first congruence, the rest follows by identifying which angles and sides correspond between the smaller triangles.

Question 13. In the adjoining figure, AP ⊥ l and PR > PQ. Show that AR > AQ.
Answer: Mark a point S on PR where PS equals PQ. Connect A to S and label the angles as indicated in the figure. Since PQ = PS, AP = AP (common side), and ∠APQ = ∠APS (both measure 90°), we have △APQ ≅ △APS by the SAS axiom. From the congruence of these triangles, their matching parts are equal, so ∠1 = ∠2. In △ARS, the exterior angle ∠2 exceeds the non-adjacent interior angle ∠3, which gives us ∠1 > ∠3. Since the side opposite a larger angle is always longer, AR > AQ.
In simple words: We create a new triangle that is exactly the same shape and size as another triangle. This lets us compare angles. A bigger angle in a triangle always has a longer side opposite to it, so AR must be longer than AQ.

Exam Tip: The key step is recognizing that constructing point S allows you to use triangle congruence (SAS). Then apply the exterior angle theorem and the rule that larger angles face longer sides.

Question 14. If O is any point in the interior of a triangle ABC, show that OA + OB + OC > ½(AB + BC + CA).
Answer: Using the triangle inequality theorem, which states that the sum of any two sides of a triangle exceeds the third side, we establish three inequalities. In △OBC, we have OB + OC > BC ......(i). Similarly, in △OCA, we obtain OC + OA > CA .......(ii). And in △OAB, we get OA + OB > AB .......(iii). When we add all three inequalities together, the left side becomes OB + OC + OC + OA + OA + OB, which simplifies to 2(OA + OB + OC). The right side is BC + CA + AB. Therefore, 2(OA + OB + OC) > AB + BC + CA. Dividing both sides by 2 gives us OA + OB + OC > ½(AB + BC + CA).
In simple words: If you pick any point inside a triangle and measure its distance to all three corners, those three distances added together will always be more than half the perimeter of the triangle.

Exam Tip: Make sure to write out all three triangle inequality statements clearly, then add them carefully. Dividing by 2 at the end is the crucial final step that completes the proof.