Maharashtra Board Class 10 Maths Chapter 1 Similarity Set 1.2 Solutions

Question 1. Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में एक त्रिभुज PQR दिखाया गया है जिसमें बिंदु M भुजा QR पर है, रेखाखंड PM कोण QPR को संभावित रूप से समद्विभाजित करता है। दूसरे चित्र में एक त्रिभुज PQR है जहाँ M भुजा QR पर है, रेखाखंड PM कोण QPR को संभावित रूप से समद्विभाजित नहीं करता है। तीसरे चित्र में एक त्रिभुज PQR है जहाँ M भुजा QR पर है, रेखाखंड PM कोण QPR को संभावित रूप से समद्विभाजित करता है।
Answer: Solution: In Δ PQR, \(\frac{PQ}{PR} = \frac{7}{3}\) (i) \(\frac{QM}{RM} = \frac{3.5}{1.5} = \frac{35}{15} = \frac{7}{3}\) (ii) \(\therefore \frac{PQ}{PR} = \frac{QM}{RM}\) [From (i) and (ii)] \(\therefore\) Ray PM is the bisector of ∠QPR. [Converse of angle bisector theorem]
(ii). In ΔPQR, \(\frac{PQ}{PR} = \frac{10}{7}\) (i) \(\frac{QM}{RM} = \frac{8}{6} = \frac{4}{3}\) (ii) \(\therefore \frac{PQ}{PR} \neq \frac{QM}{RM}\) [From (i) and (ii)] \(\therefore\) Ray PM is not the bisector of ∠QPR
(iii). In ΔPQR, \(\frac{PQ}{PR} = \frac{9}{10}\) (i) \(\frac{QM}{RM} = \frac{3.6}{4} = \frac{36}{40} = \frac{9}{10}\) (ii) \(\therefore \frac{PQ}{PR} = \frac{QM}{RM}\) [From (i) and (ii)] \(\therefore\) Ray PM is the bisector of ∠QPR [Converse of angle bisector theorem]
In simple words: To check if a ray is an angle bisector, we use the Angle Bisector Theorem or its converse. If the ratio of the sides forming the angle equals the ratio of the segments on the opposite side, then the ray is an angle bisector. In this problem, we compare these ratios for each triangle to determine if ray PM is the bisector of ∠QPR.

🎯 Exam Tip: Remember the conditions for the Angle Bisector Theorem and its converse; correctly calculating the ratios of the sides and segments is crucial for accurate identification.

Question 2. In ΔPQR PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज PQR है। बिंदु N भुजा PR पर और बिंदु M भुजा PQ पर स्थित है। एक रेखाखंड NM खींचा गया है। दिए गए मानों के आधार पर हमें यह निर्धारित करना है कि रेखा NM भुजा RQ के समानांतर है या नहीं।
Answer: Solution: PN + NR = PR [P-N-R] \(\therefore\) PN + 8 = 20 \(\therefore\) PN = 20 - 8 = 12 Also, PM + MQ = PQ [P-M-Q] \(\therefore\) 15 + MQ = 25 \(\therefore\) MQ = 25 - 15 = 10 \(\frac{PN}{NR} = \frac{12}{8}\) \(\therefore \frac{PN}{NR} = \frac{3}{2}\) (i) \(\frac{PM}{MQ} = \frac{15}{10}\) \(\therefore \frac{PM}{MQ} = \frac{3}{2}\) (ii) In ΔPQR, \(\frac{PN}{NR} = \frac{PM}{MQ}\) [From (i) and (ii)] \(\therefore\) line NM || side RQ [Converse of basic proportionality theorem]
In simple words: To check if line NM is parallel to side RQ, we use the Converse of the Basic Proportionality Theorem. We calculate the ratios of the segments PN/NR and PM/MQ. Since both ratios are equal (3/2), line NM is parallel to side RQ.

🎯 Exam Tip: Ensure precise calculations of segment lengths and their ratios. The Converse of BPT is a fundamental concept for proving parallel lines in triangles.

Question 3. In ΔMNP, NQ is a bisector of ∠N. If MN = 5, PN = 7, MQ = 2.5, then find QP.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज MNP है जिसमें NQ कोण N का समद्विभाजक है। बिंदु Q भुजा MP पर स्थित है। भुजाओं MN, PN और MQ की लंबाई दी गई है, और हमें QP की लंबाई ज्ञात करनी है।
Answer: Solution: In ΔMNP, NQ is the bisector of ∠N. [Given] \(\therefore \frac{PN}{MN} = \frac{QP}{MQ}\) [Property of angle bisector of a triangle] \(\frac{7}{5} = \frac{QP}{2.5}\) \(\therefore QP = \frac{7 \times 2.5}{5}\) \(\therefore\) QP = 3.5 units
In simple words: We apply the Angle Bisector Theorem to triangle MNP. Since NQ bisects angle N, the ratio of the sides forming the angle (PN/MN) is equal to the ratio of the segments on the opposite side (QP/MQ). Using the given values, we can calculate QP.

🎯 Exam Tip: Clearly identify the angle bisector and the corresponding sides and segments. Algebraic manipulation should be precise to avoid calculation errors.

Question 4. Measures of some angles in the figure are given. Prove that \(\frac{AP}{PB} = \frac{AQ}{QC}\).
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज ABC है। बिंदु P भुजा AB पर और बिंदु Q भुजा AC पर स्थित है। एक रेखाखंड PQ खींचा गया है। कोण APQ और कोण ABC दोनों 60° दिए गए हैं। हमें सिद्ध करना है कि भुजाओं का अनुपात AP/PB, AQ/QC के बराबर है।
Answer: Solution: Proof ∠APQ = ∠ABC = 60° [Given] \(\therefore\) ∠APQ = ∠ABC \(\therefore\) side PQ || side BC (i) [Corresponding angles test] In ΔABC, sidePQ || sideBC [From (i)] \(\therefore \frac{AP}{PB} = \frac{AQ}{QC}\) [Basic proportionality theorem]
In simple words: Since corresponding angles ∠APQ and ∠ABC are equal (60°), line PQ is parallel to line BC. By the Basic Proportionality Theorem (BPT), if a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally. This proves the required ratio.

🎯 Exam Tip: Recognize the conditions for parallel lines (e.g., corresponding angles) and then apply the Basic Proportionality Theorem correctly. Clear step-by-step reasoning is essential.

Question 5. In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक समलंब ABCD है जहाँ AB, PQ और DC तीनों भुजाएँ समानांतर हैं। बिंदु P भुजा AD पर और बिंदु Q भुजा BC पर स्थित है। हमें AP, PD और QC की लंबाई दी गई है, और हमें BQ की लंबाई ज्ञात करनी है।
Answer: Solution: side AB || side PQ || side DC [Given] \(\therefore \frac{AP}{PD} = \frac{BQ}{QC}\) [Property of three parallel lines and their transversals] \(\frac{15}{12} = \frac{BQ}{14}\) \(\therefore BQ = \frac{15 \times 14}{12}\) \(\therefore\) BQ = 17.5 units
In simple words: When three parallel lines intersect two transversals, they cut off proportional segments on the transversals. Using the given lengths and this property, we can set up a proportion and solve for the unknown length BQ.

🎯 Exam Tip: This problem utilizes the property of intercepts made by three parallel lines on transversals. Setting up the correct ratio is the key to finding the unknown length. Ensure accurate arithmetic.

Question 6. Find QP using given information in the figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज MNP है जिसमें NQ कोण N को समद्विभाजित करता हुआ एक रेखाखंड है। बिंदु Q भुजा MP पर स्थित है। भुजा MN की लंबाई 25, भुजा NP की लंबाई 40 और भुजा MQ की लंबाई 14 दी गई है। हमें QP की लंबाई ज्ञात करनी है।
Answer: Solution: In ΔMNP, seg NQ bisects ∠N. [Given] \(\therefore \frac{PN}{MN} = \frac{QP}{MQ}\) [Property of angle bisector of a triangle] \(\therefore \frac{40}{25} = \frac{QP}{14}\) \(\therefore QP = \frac{40 \times 14}{25}\) \(\therefore\) QP = 22.4 units
In simple words: The Angle Bisector Theorem states that if a ray bisects an angle of a triangle, it divides the opposite side into two segments proportional to the other two sides. Applying this to ΔMNP with NQ as the bisector, we set up the ratio PN/MN = QP/MQ and solve for QP.

🎯 Exam Tip: Double-check the labeling of the triangle and the corresponding ratios according to the Angle Bisector Theorem. Calculation of the final value must be precise.

Question 7. In the adjoining figure, if AB || CD || FE, then find x and AE.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में तीन समानांतर रेखाएँ AB, CD और FE हैं जिन्हें दो तिर्यक रेखाएँ काट रही हैं। तिर्यक रेखा के एक तरफ के खंड BD और DF की लंबाई क्रमशः 8 और 4 है। दूसरी तिर्यक रेखा के खंड AC और CE की लंबाई क्रमशः 12 और x है। हमें x और AE का मान ज्ञात करना है।
Answer: Solution: line AB || line CD || line FE [Given] \(\therefore \frac{BD}{DF} = \frac{AC}{CE}\) [Property of three parallel lines and their transversals] \(\frac{8}{4} = \frac{12}{X}\) \(\therefore X = \frac{12 \times 4}{8}\) \(\therefore\) X = 6 units Now, AE = AC + CE [A-C-E] = 12 + x = 12 + 6 = 18 units \(\therefore\) x = 6 units and AE = 18 units
In simple words: Since three parallel lines cut two transversals, the segments intercepted on the transversals are proportional. We use this property to find x, and then add AC and CE to find the total length AE.

🎯 Exam Tip: Clearly identify the parallel lines and transversals. The property of proportional intercepts is key. Remember to calculate both 'x' and 'AE' as requested.

Question 8. In ΔLMN, ray MT bisects ∠LMN. If LM = 6, MN = 10, TN = 8, then find LT.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज LMN है जहाँ ray MT कोण LMN को समद्विभाजित करती है। बिंदु T भुजा LN पर स्थित है। भुजाओं LM, MN और TN की लंबाई दी गई है, और हमें LT की लंबाई ज्ञात करनी है।
Answer: Solution: In ΔLMN, ray MT bisects ∠LMN. [Given] \(\therefore \frac{LM}{MN} = \frac{LT}{TN}\) [Property of angle bisector of a triangle] \(\frac{6}{10} = \frac{LT}{8}\) \(\therefore LT = \frac{6 \times 8}{10}\) \(\therefore\) LT = 4.8 units
In simple words: By the Angle Bisector Theorem, if ray MT bisects angle LMN, then the ratio of the sides LM/MN is equal to the ratio of the segments LT/TN. We use the given values to solve for LT.

🎯 Exam Tip: Accurately apply the Angle Bisector Theorem by matching the correct sides to the correct segments. Pay attention to the labels of the triangle vertices and segments for precise calculation.

Question 9. In ΔABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x - 2, DC = x + 2, then find the value of x.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज ABC है जहाँ रेखाखंड BD कोण ABC को समद्विभाजित करता है। बिंदु D भुजा AC पर स्थित है। भुजाओं AB, BC, AD और DC की लंबाई x के पदों में दी गई है। हमें x का मान ज्ञात करना है।
Answer: Solution: In ΔABC, seg BD bisects ∠ABC. [Given] \(\therefore \frac{AB}{BC} = \frac{AD}{CD}\) [Property of angle bisector of a triangle] \(\therefore \frac{x}{x+5} = \frac{x-2}{x+2}\) \(\therefore x(x + 2) = (x - 2)(x + 5)\) \(\therefore x^2 + 2x = x^2 + 5x - 2x - 10\) \(\therefore 2x = 3x - 10\) \(\therefore 10 = 3x - 2x\) \(\therefore\) x = 10
In simple words: Using the Angle Bisector Theorem, we set up a proportion relating the sides AB/BC to the segments AD/DC. Substituting the given expressions in terms of x, we form an algebraic equation. Solving this equation yields the value of x.

🎯 Exam Tip: Carefully set up the proportion using the Angle Bisector Theorem. Algebraic simplification, especially involving quadratic terms, needs precision to find the correct value of x.

Question 10. In the adjoining figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज DEF है। त्रिभुज के अंदर एक बिंदु X है जिसे D, E और F से जोड़ा गया है। बिंदु P रेखाखंड XD पर, Q रेखाखंड XE पर और R रेखाखंड XF पर स्थित है। रेखाखंड PQ रेखाखंड DE के समानांतर है, और रेखाखंड QR रेखाखंड EF के समानांतर है। हमें सिद्ध करना है कि रेखाखंड PR रेखाखंड DF के समानांतर है।
Answer: Solution: In ΔXDE, PQ || DE \(\frac{XP}{PD} = \frac{XQ}{QE}\) (i) [Basic proportionality theorem] In ΔXEF, QR || EF \(\frac{XR}{RF} = \frac{XQ}{QE}\) (ii) [Basic proportionality theorem] \(\therefore \frac{XP}{PD} = \frac{XR}{RF}\) [From (i) and (ii)] \(\therefore\) seg PR || seg DF [Converse of basic proportionality theorem]
In simple words: We apply the Basic Proportionality Theorem (BPT) twice. First, in triangle XDE, since PQ || DE, we get XP/PD = XQ/QE. Second, in triangle XEF, since QR || EF, we get XR/RF = XQ/QE. By combining these two results, we find that XP/PD = XR/RF. This condition, by the Converse of BPT in triangle XDF, implies that PR || DF.

🎯 Exam Tip: This problem requires a sequential application of BPT and its converse. Ensure accurate identification of the triangles and parallel lines in each step to establish the proportionalities correctly.

Question 11. In ΔABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज ABC है। ray BD कोण ABC को समद्विभाजित करती है और ray CE कोण ACB को समद्विभाजित करती है। बिंदु D भुजा AC पर और बिंदु E भुजा AB पर स्थित है। दिया गया है कि भुजा AB, भुजा AC के बराबर है। हमें सिद्ध करना है कि रेखाखंड ED, रेखाखंड BC के समानांतर है।
Answer: Solution: In ΔABC, ray BD bisects ∠ABC. [Given] \(\therefore \frac{AB}{BC} = \frac{AD}{DC}\) (i) [Property of angle bisector of a triangle] Also, in ΔABC, ray CE bisects ∠ACB. [Given] \(\therefore \frac{AC}{BC} = \frac{AE}{EB}\) (ii) [Property of angle bisector of a triangle] But, seg AB = seg AC (iii) [Given] \(\therefore \frac{AB}{BC} = \frac{AE}{EB}\) (iv) [From (ii) and (iii)] \(\therefore \frac{AD}{DC} = \frac{AE}{EB}\) [From (i) and (iv)] \(\therefore\) ED || BC [Converse of basic proportionality theorem]
In simple words: We apply the Angle Bisector Theorem to both bisectors, BD and CE, creating two sets of proportional ratios. Given that AB = AC, we can equate these ratios to show that AD/DC = AE/EB. By the Converse of the Basic Proportionality Theorem, this proportionality implies that line segment ED is parallel to BC.

🎯 Exam Tip: This proof involves multiple theorems. Systematically apply the Angle Bisector Theorem for both bisectors, then use the given equality (AB=AC) to relate the ratios, and finally use the Converse of BPT to conclude parallelism.

Question 1. i. Draw a ΔABC. ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D. iii. Measure the sides. AB = 4 cm, BC = 4 cm, AD = 2 cm, DC = 2 cm iv. Find ratios \(\frac{AB}{BC}\) and \(\frac{AD}{DC}\) v. You will find that both the ratios are almost equal. vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज ABC दर्शाया गया है। कोण B को एक रेखाखंड BD द्वारा समद्विभाजित किया गया है, जो भुजा AC को बिंदु D पर काटता है। भुजाओं AB, BC, AD और DC की लंबाई के मान दिए गए हैं ताकि छात्रों को कोण समद्विभाजक प्रमेय के गुणों को सत्यापित करने में मदद मिल सके।
Answer: Solution: AB = 4 cm, BC = 4 cm AD = 2 cm, DC = 2 cm \(\frac{AB}{BC} = \frac{4}{4} = 1\) (i) \(\frac{AD}{DC} = \frac{2}{2} = 1\) (ii) \(\therefore \frac{AB}{BC} = \frac{AD}{DC}\) [From (i) and (ii)] Note: Students should bisect the remaining angles and verify that the ratios are equal.
In simple words: This activity asks students to draw a triangle, bisect an angle, measure the sides and segments created, and then calculate the ratios AB/BC and AD/DC. The measurements provided illustrate the Angle Bisector Theorem, showing that these ratios are equal. Students are encouraged to repeat the process for other angles to confirm the theorem.

🎯 Exam Tip: For practical geometry questions, accuracy in drawing and measuring is paramount. Understanding the theoretical concept (Angle Bisector Theorem) makes it easier to predict and verify the ratios.

Question 2. Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof. i. The areas of two triangles of equal height are proportional to their bases. ii. Every point on the bisector of an angle is equidistant from the sides of the angle. (Textbook pg. no. 9)
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज CAB है जिसमें AD कोण A का समद्विभाजक है। बिंदु D भुजा CB पर स्थित है। रचना के लिए, DM भुजा AB पर लंबवत है और DN भुजा AC पर लंबवत है। यह चित्र कोण समद्विभाजक प्रमेय के एक वैकल्पिक प्रमाण को दर्शाता है।
Answer: Solution: Given: In ΔCAB, ray AD bisects ∠A. To prove: \(\frac{AB}{AC} = \frac{BD}{DC}\) Construction: Draw seg DM \(\perp\) seg AB A-M-B and seg DN \(\perp\) seg AC, A-N-C. Proof: In ΔABC, Point D is on angle bisector of ∠A. [Given] \(\therefore\) DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle] \(\frac{A(\Delta ABD)}{A(\Delta ACD)} = \frac{AB \times DM}{AC \times DN}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights] \(\therefore \frac{A(\Delta ABD)}{A(\Delta ACD)} = \frac{AB}{AC}\) (ii) [From (i)] Also, ΔABD and ΔACD have equal height. \(\frac{A(\Delta ABD)}{A(\Delta ACD)} = \frac{BD}{CD}\) (iii) [Triangles having equal height] \(\therefore \frac{AB}{AC} = \frac{BD}{CD}\) [From (ii) and (iii)]
In simple words: This proof uses the property that any point on an angle bisector is equidistant from the sides of the angle (DM=DN) and the area ratio of triangles. By comparing the area ratios of ΔABD and ΔACD based on their bases and heights, and also based on their common height from A to BC, we establish the proportionality AB/AC = BD/DC.

🎯 Exam Tip: This alternative proof highlights the connection between angle bisectors, perpendicular distances, and areas of triangles. Clearly state each property used and ensure the logical flow of the argument.

Question 3. i. Draw three parallel lines. ii. Label them as l, m, n. iii. Draw transversals t\(_1\) and t\(_2\). iv. AB and BC are intercepts on transversal t\(_1\). v. PQ and QR are intercepts on transversal t\(_2\). vi. Find ratios \(\frac{AB}{BC}\) and \(\frac{PQ}{QR}\) You will find that they are almost equal. Verify that they are equal.(Textbook pg, no. 10)
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में तीन समानांतर रेखाएँ l, m, n हैं। इन्हें दो तिर्यक रेखाएँ t1 और t2 काट रही हैं। तिर्यक रेखा t1 पर खंड AB और BC हैं, जबकि तिर्यक रेखा t2 पर खंड PQ और QR हैं। उदाहरण के लिए दी गई भुजाओं की लंबाई का उपयोग करके, हमें यह सत्यापित करना है कि इन खंडों का अनुपात (AB/BC और PQ/QR) समान है।
Answer: Solution: Here, AB = 1.5 cm, BC = 2.1 cm, PQ = 1.7 cm, QR = 2.3 cm \(\frac{AB}{BC} = \frac{1.5}{2.1} = 0.714 \approx 0.7\) \(\frac{PQ}{QR} = \frac{1.7}{2.3} = 0.739 \approx 0.7\) \(\therefore \frac{AB}{BC} = \frac{PQ}{QR}\) (Students should draw figures similar to the ones given and verify the properties.)
In simple words: This activity demonstrates the property of intercepts made by three parallel lines. By drawing parallel lines and transversals, then measuring the segments, students can confirm that the ratios of corresponding segments on the transversals are equal, which is a key geometric property.

🎯 Exam Tip: This exercise reinforces the understanding of the theorem concerning three parallel lines and their transversals. Practical measurement and calculation of ratios are essential for verification.

Question 4. In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF. (Textbook pg, no. 12)
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में तीन समानांतर रेखाएँ AB, CD और EF हैं। ये रेखाएँ दो तिर्यक रेखाओं को काटती हैं। एक तिर्यक रेखा पर AC और CE खंड हैं, जिनकी लंबाई क्रमशः 5.4 और 9 है। दूसरी तिर्यक रेखा पर BD और DF खंड हैं, जिनमें BD की लंबाई 7.5 है। हमें DF की लंबाई ज्ञात करनी है।
Answer: Solution: AB || CD || EF [Given] \(\therefore \frac{AC}{CE} = \frac{BD}{DF}\) [Property of three parallel lines and their transversals] \(\frac{5.4}{9} = \frac{7.5}{DF}\) \(\therefore DF = \frac{7.5 \times 9}{5.4}\) \(\therefore\) DF = 12.5 units
In simple words: According to the property of three parallel lines and their transversals, the ratios of the intercepted segments on the transversals are equal. We set up the proportion AC/CE = BD/DF with the given values to solve for DF.

🎯 Exam Tip: Correctly identify the corresponding segments formed by the parallel lines on the transversals. Precision in setting up the ratio and performing the calculation is vital for finding the unknown length DF.

Question 5. In ΔABC, ray BD bisects ∠ABC. A-D-C, side DE || side BC, A-E-B, then prove that \(\frac{AB}{BC} = \frac{AE}{EB}\). (Textbook pg, no. 13)
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज ABC है। ray BD कोण ABC को समद्विभाजित करती है, और बिंदु D भुजा AC पर है। एक रेखाखंड DE भुजा BC के समानांतर खींचा गया है, जहाँ बिंदु E भुजा AB पर है। हमें यह सिद्ध करना है कि AB/BC = AE/EB।
Answer: Solution: In ΔABC, ray BD bisects ∠B. [Given] \(\therefore \frac{AB}{BC} = \frac{AD}{DC}\) (i) [Angle bisector theorem] In ΔABC, DE || BC [Given] \(\therefore \frac{AE}{EB} = \frac{AD}{DC}\) (ii) [Basic proportionality theorem] \(\therefore \frac{AB}{BC} = \frac{AE}{EB}\) [From (i) and (ii)]
In simple words: We first use the Angle Bisector Theorem for ray BD to establish the ratio AB/BC = AD/DC. Then, since DE || BC, we apply the Basic Proportionality Theorem to get AE/EB = AD/DC. By comparing these two results, we conclude that AB/BC = AE/EB.

🎯 Exam Tip: This proof combines the Angle Bisector Theorem and the Basic Proportionality Theorem. Clearly state which theorem is being applied in each step and how the derived ratios lead to the final proof.