NCERT Solutions Class 9 Mathematics Chapter 8 Quadrilaterals

Exercise 8.1

Q.1) The angles of a quadrilateral are in the ratio 3 ∶ 5 ∶ 9 ∶ 13. Find all the angles of the quadrilateral.
Sol.1) Suppose the measures of four angles are 3𝑥, 5𝑥, 9𝑥 and 13𝑥.
∴ 3𝑥 + 5𝑥 + 9𝑥 + 13𝑥 = 360° [Angle sum property of a quadrilateral]
⇒ 30𝑥 = 360°
⇒ 𝑥 = 360/30 = 12°
⇒ 3𝑥 = 3 × 12° = 36°
5𝑥 = 5 × 12° = 60°
9𝑥 = 9 × 12° = 108°
13𝑥 = 13 × 12° = 156°
∴ the angles of the quadrilateral are 36°, 60°, 108° and 156° Ans.

Q.2) If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Sol.2) Given : ABCD is a parallelogram in which 𝐴𝐶 = 𝐵𝐷.
To Prove : ABCD is a rectangle.
Proof : In Δ𝐴𝐵𝐶 and Δ𝐴𝐵𝐷
𝐴𝐵 = 𝐴𝐵                   [Common]
𝐵𝐶 = 𝐴𝐷                   [Opposite sides of a parallelogram]
𝐴𝐶 = 𝐵𝐷                   [Given]
∴ Δ𝐴𝐵𝐶 ≅ Δ𝐵𝐴𝐷      [SSS congruence]
∠𝐴𝐵𝐶 = ∠𝐵𝐴𝐷 ...(i)         [CPCT]
Since, ABCD is a parallelogram, thus,
∠𝐴𝐵𝐶 + ∠𝐵𝐴𝐷 = 180° ...(ii)         [Consecutive interior angles]
∠𝐴𝐵𝐶 + ∠𝐴𝐵𝐶 = 180°
∴ 2∠𝐴𝐵𝐶 = 180°                         [From (i) and (ii)]
⇒ ∠𝐴𝐵𝐶 = ∠𝐵𝐴𝐷 = 90°
This shows that ABCD is a parallelogram one of whose angle is 90°.
Hence, ABCD is a rectangle. Proved.

Q.3) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Sol.3) Given : A quadrilateral ABCD, in which diagonals AC and BD bisect each other at right angles.
To Prove : ABCD is a rhombus.
Proof : In Δ𝐴𝑂𝐵 and Δ𝐵𝑂𝐶
𝐴𝑂 = 𝑂𝐶 [Diagonals AC and BD bisect each other]
∠𝐴𝑂𝐵 = ∠𝐶𝑂𝐵         [Each = 90°]
𝐵𝑂 = 𝐵𝑂                 [Common]
∴ Δ𝐴𝑂𝐵 ≅ Δ𝐵𝑂𝐶 [SAS congruence]
𝐴𝐵 = 𝐵𝐶 ...(i)         [CPCT]
Since, ABCD is a quadrilateral in which
𝐴𝐵 = 𝐵𝐶                [From (i)]
Hence, ABCD is a rhombus.
[∵ if the diagonals of a quadrilateral bisect each other, then it is a parallelogram and
opposite sides of a parallelogram are equal]      Proved.

Q.4) Show that the diagonals of a square are equal and bisect each other at right angles.
Sol.4) Given : ABCD is a square in which AC and BD are diagonals.
To Prove : 𝐴𝐶 = 𝐵𝐷 and AC bisects BD at right angles, i.e. 𝐴𝐶 ⊥ 𝐵𝐷.
𝐴𝑂 = 𝑂𝐶, 𝑂𝐵 = 𝑂𝐷
Proof : In Δ𝐴𝐵𝐶 and Δ𝐵𝐴𝐷,
𝐴𝐵 = 𝐴𝐵           [Common]
𝐵𝐶 = 𝐴𝐷           [Sides of a square]
∠𝐴𝐵𝐶 = ∠𝐵𝐴𝐷 = 90°           [Angles of a square]
∴ Δ𝐴𝐵𝐶 ≅ Δ𝐵𝐴𝐷      [SAS congruence]
⇒ 𝐴𝐶 = 𝐵𝐷             [CPCT]
Now in Δ𝐴𝑂𝐵 and Δ𝐶𝑂𝐷,
𝐴𝐵 = 𝐷𝐶                     [Sides of a square]
∠𝐴𝑂𝐵 = ∠𝐶𝑂𝐷             [Vertically opposite angles]
∠𝑂𝐴𝐵 = ∠𝑂𝐶𝐷                        [Alternate angles]
∴ Δ𝐴𝑂𝐵 ≅ Δ𝐶𝑂𝐷                     [AAS congruence]
∠𝐴𝑂 = ∠𝑂𝐶                     [CPCT]
Similarly by taking Δ𝐴𝑂𝐷 and Δ𝐵𝑂𝐶, we can show that 𝑂𝐵 = 𝑂𝐷.
In Δ𝐴𝐵𝐶, ∠𝐵𝐴𝐶 + ∠𝐵𝐶𝐴 = 90°           [ ∠𝐵 = 90°]
⇒ 2∠𝐵𝐴𝐶 = 90°           [∠𝐵𝐴𝐶 = ∠𝐵𝐶𝐴, 𝑎𝑠 𝐵𝐶 = 𝐴𝐷]
⇒ ∠𝐵𝐶𝐴 = 45° 𝑜𝑟 ∠𝐵𝐶𝑂 = 45°
Similarly ∠𝐶𝐵𝑂 = 45°
In Δ𝐵𝐶𝑂.
∠𝐵𝐶𝑂 + ∠𝐶𝐵𝑂 + ∠𝐵𝑂𝐶 = 180°
⇒ 90° + ∠𝐵𝑂𝐶 = 180°
⇒ ∠𝐵𝑂𝐶 = 90°
⇒ 𝐵𝑂 ⊥ 𝑂𝐶 ⇒ 𝐵𝑂 ⊥ 𝐴𝐶
Hence, 𝐴𝐶 = 𝐵𝐷, 𝐴𝐶 ⊥ 𝐵𝐷, 𝐴𝑂 = 𝑂𝐶 and 𝑂𝐵 = 𝑂𝐷. Proved.

Q.5) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Sol.5) Given : A quadrilateral ABCD, in which
diagonals AC and BD are equal and bisect
each other at right angles,
To Prove : ABCD is a square.
Proof : Since ABCD is a quadrilateral whose diagonals bisect each other, so it is a parallelogram. Also, its diagonals bisect each other at right angles, therefore, ABCD is a rhombus.
⇒ 𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 = 𝐷𝐴 [Sides of a rhombus]
In Δ𝐴𝐵𝐶 and Δ𝐵𝐴𝐷, we have
𝐴𝐵 = 𝐴𝐵           [Common]
𝐵𝐶 = 𝐴𝐷           [Sides of a rhombus]
𝐴𝐶 = 𝐵𝐷           [Given]
∴ Δ𝐴𝐵𝐶 ≅ Δ𝐵𝐴𝐷           [SSS congruence]
∴ ∠𝐴𝐵𝐶 = ∠𝐵𝐴𝐷           [CPCT]
But, ∠𝐴𝐵𝐶 + ∠𝐵𝐴𝐷 = 180°            [Consecutive interior angles]
∠𝐴𝐵𝐶 = ∠𝐵𝐴𝐷 = 90°
∠𝐴 = ∠𝐵 = ∠𝐶 = ∠𝐷 = 90°              [Opposite angles of a ||𝑔𝑚]

⇒ ABCD is a rhombus whose angles are of 90° each.
Hence, ABCD is a square. Proved.

Q.6) Diagonal AC of a parallelogram 𝐴𝐵𝐶𝐷 bisects ∠𝐴 (see Fig.). Show that
(i) it bisects ∠𝐶 also,
(ii) 𝐴𝐵𝐶𝐷 is a rhombus.
Sol.6) To Prove :
(i) Diagonal AC bisects ∠𝐶 i.e., ∠𝐷𝐶𝐴 = ∠𝐵𝐶𝐴
(ii) ABCD is a rhombus.
Proof :
(i) ∠𝐷𝐴𝐶 = ∠𝐵𝐶𝐴              [Alternate angles]
∠𝐵𝐴𝐶 = ∠𝐷𝐶𝐴                  [Alternate angles]
But, ∠𝐷𝐴𝐶 = ∠𝐵𝐴𝐶           [Given]
∴ ∠𝐵𝐶𝐴 = ∠𝐷𝐶𝐴
Hence, 𝐴𝐶 bisects ∠𝐷𝐶𝐵
Or, 𝐴𝐶 bisects ∠𝐶 Proved.
(ii) In Δ𝐴𝐵𝐶 and Δ𝐶𝐷𝐴
𝐴𝐶 = 𝐴𝐶                 [Common]
∠𝐵𝐴𝐶 = ∠𝐷𝐴𝐶         [Given]
and ∠𝐵𝐶𝐴 = ∠𝐷𝐴𝐶             [Proved above]
∴ Δ𝐴𝐵𝐶 ≅ Δ𝐴𝐷𝐶                 [ASA congruence]
∴ 𝐵𝐶 = 𝐷𝐶                    [CPCT]
But 𝐴𝐵 = 𝐷𝐶                 [Given]
∴ 𝐴𝐵 = 𝐵𝐶 = 𝐷𝐶 = 𝐴𝐷                 [∵ opposite angles are equal]
Hence, ABCD is a rhombus Proved.

Q.7) ABCD is a rhombus. Show that diagonal 𝐴𝐶 bisects ∠ 𝐴 as well as ∠𝐶 and diagonal BD bisects ∠𝐵 as well as ∠𝐷.
Sol.7) Given : ABCD is a rhombus, i.e.,
𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 = 𝐷𝐴.
To Prove : ∠𝐷𝐴𝐶 = ∠𝐵𝐴𝐶,
∠𝐵𝐶𝐴 = ∠𝐷𝐶𝐴
∠𝐴𝐷𝐵 = ∠𝐶𝐷𝐵, ∠𝐴𝐵𝐷 = ∠𝐶𝐵𝐷
Proof : In Δ𝐴𝐵𝐶 and Δ𝐶𝐷𝐴, we have
𝐴𝐵 = 𝐴𝐷         [Sides of a rhombus]
𝐴𝐶 = 𝐴𝐶         [Common]
𝐵𝐶 = 𝐶𝐷         [Sides of a rhombus]
Δ𝐴𝐵𝐶 ≅ Δ𝐴𝐷𝐶 [SSS congruence]
So, ∠𝐷𝐴𝐶 = ∠𝐵𝐴𝐶         [CPCT]
∠𝐵𝐶𝐴 = ∠𝐷𝐶𝐴               [CPCT]
Similarly, ∠𝐴𝐷𝐵 = ∠𝐶𝐷𝐵 and ∠𝐴𝐵𝐷 = ∠𝐶𝐵𝐷.
Hence, diagonal AC bisects ∠𝐴 as well as ∠𝐶 and diagonal BD bisects ∠𝐵 as well as ∠𝐷. Proved.

Q.8) ABCD is a rectangle in which diagonal AC bisects ∠ 𝐴 as well as ∠𝐶. Show that :
(i) ABCD is a square (ii) diagonal BD bisects ∠𝐵 as well as ∠𝐷.
Sol.8) Given : ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠𝐶.
To Prove : (i) ABCD is a square.
(ii) Diagonal BD bisects ∠𝐵 as well as ∠𝐷.
Proof : (i) In Δ𝐴𝐵𝐶 and Δ𝐴𝐷𝐶, we have
∠𝐵𝐴𝐶 = ∠𝐷𝐴𝐶                [Given]
∠𝐵𝐶𝐴 = ∠𝐷𝐶𝐴                [Given]
𝐴𝐶 = 𝐴𝐶
∴ Δ𝐴𝐵𝐶 ≅ Δ𝐴𝐷𝐶 [ASA congruence]
∴ 𝐴𝐵 = 𝐴𝐷 𝑎𝑛𝑑 𝐶𝐵 = 𝐶𝐷              [CPCT]
But, in a rectangle opposite sides are equal, i.e., 𝐴𝐵 = 𝐷𝐶 and 𝐵𝐶 = 𝐴𝐷
∴ 𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 = 𝐷𝐴
Hence, ABCD is a square Proved.
(ii) In Δ𝐴𝐵𝐷 and Δ𝐶𝐷𝐵, we have
𝐴𝐷 = 𝐶𝐷
𝐴𝐵 = 𝐶𝐷                           [Sides of a square]
𝐵𝐷 = 𝐵𝐷                           [Common]
∴ Δ𝐴𝐵𝐷 ≅ Δ𝐶𝐵𝐷                 [SSS congruence]
So, ∠𝐴𝐵𝐷 = ∠𝐶𝐵𝐷
∠𝐴𝐷𝐵 = ∠𝐶𝐷𝐵
Hence, diagonal BD bisects ∠𝐵 as well as ∠𝐷 Proved.

Q.9) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig.). Show that :
(i) Δ 𝐴𝑃𝐷 ≅ Δ𝐶𝑄𝐵
(ii) 𝐴𝑃 = 𝐶𝑄
(iii) Δ 𝐴𝑄𝐵 ≅ Δ𝐶𝑃𝐷
(iv) 𝐴𝑄 = 𝐶𝑃
(𝑣) 𝐴𝑃𝐶𝑄 is a parallelogram

Sol.9) Given : ABCD is a parallelogram and P and Q are
points on diagonal BD such that 𝐷𝑃 = 𝐵𝑄.
To Prove : (i) Δ𝐴𝑃𝐷 ≅ Δ𝐶𝑄𝐵
(ii) 𝐴𝑃 = 𝐶𝑄
(iii) Δ𝐴𝑄𝐵 ≅ Δ𝐶𝑃𝐷
(iv) 𝐴𝑄 = 𝐶𝑃
(𝑣) 𝐴𝑃𝐶𝑄 is a parallelogram
Proof : (i) In Δ𝐴𝑃𝐷 and Δ𝐶𝑄𝐵, we have
𝐴𝐷 = 𝐵𝐶              [Opposite sides of a ||𝑔𝑚]
𝐷𝑃 = 𝐵𝑄              [Given]
∠ADP = ∠CBQ                [Alternate angles]
∴ Δ𝐴𝑃𝐷 ≅ Δ𝐶𝑄𝐵              [SAS congruence]
(ii) ∴ 𝐴𝑃 = 𝐶𝑄                 [CPCT]
(iii) In Δ𝐴𝑄𝐵 and Δ𝐶𝑃𝐷, we have
𝐴𝐵 = 𝐶𝐷                     [Opposite sides of a ||gm]
𝐷𝑃 = 𝐵𝑄                     [Given]
∠𝐴𝐵𝑄 = ∠𝐶𝐷𝑃              [Alternate angles]
∴ Δ𝐴𝑄𝐵 ≅ Δ𝐶𝑃𝐷              [SAS congruence]
(iv) ∴ 𝐴𝑄 = 𝐶𝑃              [CPCT]
(v) Since in 𝐴𝑃𝐶𝑄, opposite sides are equal, therefore it is a parallelogram. Proved.

Q.10) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig.). Show that
(i) Δ 𝐴𝑃𝐵 ≅ Δ𝐶𝑄𝐷
(ii) 𝐴𝑃 = 𝐶𝑄
Sol.10) Given : ABCD is a parallelogram and AP
and CQ are perpendiculars from vertices A
and C on BD.
To Prove : (i) Δ𝐴𝑃𝐵 ≅ Δ𝐶𝑄𝐷
(ii) 𝐴𝑃 = 𝐶𝑄
Proof : (i) In Δ𝐴𝑃𝐵 and Δ𝐶𝑄𝐷, we have
∠𝐴𝐵𝑃 = ∠𝐶𝐷𝑄            [Alternate angles]
𝐴𝐵 = 𝐶𝐷                    [Opposite sides of a parallelogram]
∠𝐴𝑃𝐵 = ∠𝐶𝑄𝐷            [Each = 90°]
∴ Δ𝐴𝑃𝐵 ≅ Δ𝐶𝑄𝐷           [ASA congruence]
(ii) So, 𝐴𝑃 = 𝐶𝑄          [CPCT] Proved.

Q.11) In Δ 𝐴𝐵𝐶 and Δ 𝐷𝐸𝐹, 𝐴𝐵 = 𝐷𝐸, 𝐴𝐵 || 𝐷𝐸, 𝐵𝐶 = 𝐸𝐹 and 𝐵𝐶 || 𝐸𝐹. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig.). Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilataeral BEFC is a parallelogram
(iii) 𝐴𝐷 || 𝐶𝐹 and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) 𝐴𝐶 = 𝐷𝐹
(vi) Δ 𝐴𝐵𝐶 ≡ Δ 𝐷𝐸𝐹
Sol.11) Given : In DABC and DDEF, AB = DE,
𝐴𝐵 ||𝐷𝐸, 𝐵𝐶 = 𝐸𝐹 𝑎𝑛𝑑 𝐵𝐶 || 𝐸𝐹. Vertices A, B and C are joined to vertices D, E and F.
To Prove : (i) ABED is a parallelogram
(ii) BEFC is a parallelogram
(iii) 𝐴𝐷 || 𝐶𝐹 and 𝐴𝐷 = 𝐶𝐹
(iv) ACFD is a parallelogram
(v) 𝐴𝐶 = 𝐷𝐹
(vi) Δ𝐴𝐵𝐶 ≅ Δ𝐷𝐸𝐹

Proof : (i) In quadrilateral 𝐴𝐵𝐸𝐷, we have
𝐴𝐵 = 𝐷𝐸 and 𝐴𝐵 || 𝐷𝐸.      [Given]
⇒ 𝐴𝐵𝐸𝐷 is a parallelogram.          [One pair of opposite sides is parallel and equal]
(ii) In quadrilateral 𝐵𝐸𝐹𝐶, we have
𝐵𝐶 = 𝐸𝐹 and 𝐵𝐶 || 𝐸𝐹       [Given]
⇒ 𝐵𝐸𝐹𝐶 is a parallelogram.        [One pair of opposite sides is parallel and equal]
(iii) 𝐵𝐸 = 𝐶F and 𝐵𝐸||𝐵𝐸𝐶𝐹         [BEFC is parallelogram]
𝐴𝐷 = 𝐵𝐸 and 𝐴𝐷||𝐵𝐸                  [ABED is a parallelogram]
⇒ 𝐴𝐷 = 𝐶𝐹 𝑎𝑛𝑑 𝐴𝐷||𝐶𝐹         
(iv) ACFD is a parallelogram.     [One pair of opposite sides is parallel and equal]
(v) 𝐴𝐶 = 𝐷𝐹                                 [Opposite sides of parallelogram ACFD]
(vi) In Δ𝐴𝐵𝐶 and Δ𝐷𝐸𝐹, we have
𝐴𝐵 = 𝐷𝐸          [Given]
𝐵𝐶 = 𝐸𝐹          [Given]
𝐴𝐶 = 𝐷𝐹 [Proved above]
∴ Δ𝐴𝐵𝐶 ≅ Δ𝐷𝐸𝐹 [SSS axiom] Proved.

Q.12) ABCD is a trapezium in which 𝐴𝐵||𝐶𝐷 and AD = BC (see Fig.). Show that
(i) ∠ 𝐴 = ∠ 𝐵
(ii) ∠𝐶 = ∠𝐷
(iii) Δ 𝐴𝐵𝐶 ≅ Δ 𝐵𝐴𝐷
(iv) diagonal AC = diagonal BD
Sol.12) Given : In trapezium 𝐴𝐵𝐶𝐷, 𝐴𝐵 || 𝐶𝐷 and 𝐴𝐷 = 𝐵𝐶.

To Prove : (i) ∠𝐴 = ∠𝐵
(ii) ∠𝐶 = ∠𝐷
(iii) Δ𝐴𝐵𝐶 ≅ Δ𝐵𝐴𝐷
(iv) diagonal AC = diagonal BD
Constructions : Join AC and BD. Extend AB and draw a line through C parallel to DA meeting AB produced at E.
Proof : (i) Since 𝐴𝐵 || 𝐷𝐶
⇒ 𝐴𝐸 || 𝐷𝐶 ...(i)
and 𝐴𝐷 || 𝐶𝐸 ...(ii)        [Construction]
⇒ 𝐴𝐷𝐶𝐸 is a parallelogram        [Opposite pairs of sides are parallel
∠𝐴 + ∠𝐸 = 180° ...(iii)             [Consecutive interior angles]
∠𝐵 + ∠𝐶𝐵𝐸 = 180° ...(iv)         [Linear pair]
𝐴𝐷 = 𝐶𝐸 ...(v) [Opposite sides of a ||𝑔𝑚]
𝐴𝐷 = 𝐵𝐶 ...(vi) [Given]
⇒ 𝐵𝐶 = 𝐶𝐸                              [From (v) and (vi)]
⇒ ∠𝐸 = ∠𝐶𝐵𝐸 ...(vii)               [Angles opposite to equal sides]
∴ ∠B + ∠E = 180° ...(viii)        [From (iv) and (vii)
Now from (iii) and (viii) we have
∠𝐴 + ∠𝐸 = ∠𝐵 + ∠𝐸
⇒ ∠𝐴 = ∠𝐵 Proved.

(ii) ∠𝐴 + ∠𝐷 = 180°
∠𝐵 + ∠𝐶 = 180°
⇒ ∠𝐴 + ∠𝐷 = ∠𝐵 + ∠𝐶        [∵ ∠𝐴 = ∠𝐵]
⇒ ∠𝐷 = ∠𝐶
Or ∠𝐶 = ∠𝐷 Proved.

(iii) In Δ𝐴𝐵𝐶 and Δ𝐵𝐴𝐷, we have
𝐴𝐷 = 𝐵𝐶        [Given]
∠𝐴 = ∠𝐵        [Proved]
𝐴𝐵 = 𝐶𝐷        [Common]
∴ Δ𝐴𝐵𝐶 ≅ Δ𝐵𝐴𝐷 [ASA congruence]
(iv) diagonal AC = diagonal BD         [CPCT]       Proved.

Exercise 8.2

Q.1) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. (see Fig.).
AC is a diagonal. Show that :
(i) 𝑆𝑅 || 𝐴𝐶 and 𝑆𝑅 = 1/2 𝐴𝐶
(ii) 𝑃𝑄 = 𝑆𝑅

(iii) 𝑃𝑄𝑅𝑆 is a parallelogram.
Sol.1) Given : ABCD is a quadrilateral in which P, Q, R and S are mid-points of AB, BC, CD and
DA. AC is a diagonal.
To Prove : (i) 𝑆𝑅 || 𝐴𝐶 and 𝑆𝑅 = 1/2 𝐴𝐶
(ii) 𝑃𝑄 = 𝑆𝑅
(iii) 𝑃𝑄𝑅𝑆 is a parallelogram
Proof : (i) In ΔABC, P is the mid-point of AB and Q is the mid-point of BC.
∴ 𝑃𝑄||AC and 𝑃𝑄 = 1/2 𝐴𝐶 .. (1)
In Δ𝐴𝐷𝐶, 𝑅 is the mid-point of CD and S is the mid-point of AD
∴ 𝑆𝑅||𝐴𝐶 𝑎𝑛𝑑 𝑆𝑅 = 1/2 𝐴𝐶 ..(2)                     [Mid-point theorem]
(ii) From (1) and (2), we get
𝑃𝑄 || 𝑆𝑅 and 𝑃𝑄 = 𝑆𝑅
(iii) Now in quadrilateral PQRS, its one pair of opposite sides PQ and SR is equal and parallel.
∴ 𝑃𝑄𝑅𝑆 is a parallelogram.                     Proved.

Q.2) ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Sol.2) Given : ABCD is a rhombus in which P, Q, R and S are mid points of sides AB, BC, CD and
DA respectively :
To Prove : PQRS is a rectangle.
Construction : Join AC, PR and SQ.
Proof : In ΔABC, P is mid-point of AB [Given]
Q is mid-point of BC [Given]
⇒ 𝑃𝑄 || 𝐴𝐶 and 𝑃𝑄 = 1/2 𝐴𝐶 ...(i)                     [Mid point theorem]
Similarly, in Δ𝐷𝐴𝐶, 𝑆𝑅 || 𝐴𝐶 and 𝑆𝑅 = 1/2 𝐴𝐶 ...(ii)
From (i) and (ii), we have
𝑃𝑄||𝑆𝑅 and 𝑃𝑄 = 𝑆𝑅
⇒ 𝑃𝑄𝑅𝑆 is a parallelogram                     [One pair of opposite sides is parallel and equal]
Since ABQS is a parallelogram
⇒ 𝐴𝐵 = 𝑆𝑄                     [Opposite sides of a || 𝑔𝑚]
Similarly, since PBCR is a parallelogram.
⇒ 𝐵𝐶 = 𝑃𝑅
Thus, 𝑆𝑄 = 𝑃𝑅                     [𝐴𝐵 = 𝐵𝐶]
Since SQ and PR are diagonals of parallelogram PQRS, which are equal.
⇒ 𝑃𝑄𝑅𝑆 is a rectangle.                     Proved.

Q.3) ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Sol.3) Given : A rectangle ABCD in which P, Q, R, S are the mid-points of AB, BC, CD and DA
respectively, PQ, QR, RS and SP are joined.
To Prove : PQRS is a rhombus.
Construction : Join AC.
Proof : In ΔABC, P and Q are the mid-points of the sides AB and BC.
∴ 𝑃𝑄 || 𝐴𝐶 and 𝑃𝑄 1/2 𝐴𝐶 ...(i)                     [Mid point theorem]
Similarly, in Δ𝐴𝐷𝐶,
𝑆𝑅 || 𝐴𝐶 and 𝑆𝑅 = 1/2 𝐴𝐶 ...(ii)

From (i) and (ii), we get
𝑃𝑄 || 𝑆𝑅 and 𝑃𝑄 = 𝑆𝑅 ...(iii)
Now in quadrilateral PQRS, its one pair of opposite
sides PQ and SR is parallel and equal                    [From (iii)]
∴ 𝑃𝑄𝑅𝑆 is a parallelogram.
Now 𝐴𝐷 = 𝐵𝐶 ...(iv)                    [Opposite sides of a rectangle ABCD]
∴ 1/2 𝐴𝐷 = 1/2 𝐵𝐶
⇒ 𝐴𝑆 = 𝐵𝑄
In Δ𝐴𝑃𝑆 and Δ𝐵𝑃𝑄
𝐴𝑃 = 𝐵𝑃                    [∵ P is the mid-point of AB]
𝐴𝑆 = 𝐵𝑄                    [Proved above]
∠𝑃𝐴𝑆 = ∠𝑃𝐵𝑄                    [Each = 90°]
Δ𝐴𝑃𝑆 ≅ Δ𝐵𝑃𝑄                    [SAS axiom]
∴ 𝑃𝑆 = 𝑃𝑄                    ...(v)
From (iii) and (v), we have
PQRS is a rhombus                    Proved.

Q.4) ABCD is a trapezium in which 𝐴𝐵 || 𝐷𝐶, 𝐵𝐷 is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig.). Show that F is the mid-point of BC.
Sol.4) Given :
A trapezium ABCD with AB || DC,
E is the mid-point of 𝐴𝐷 and 𝐸𝐹 || 𝐴𝐵.
To Prove : F is the mid-point of BC.
Proof : 𝐴𝐵 || 𝐷𝐶 and 𝐸𝐹 || 𝐴𝐵
⇒ 𝐴𝐵, 𝐸𝐹 and 𝐷𝐶 are parallel.
Intercepts made by parallel lines AB, EF and DC on transversal AD are equal.
∴ Intercepts made by those parallel lines on transversal BC are also equal.
i.e., 𝐵𝐹 = 𝐹𝐶
⇒ 𝐹 is the mid-point of BC.

Q.5) In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig.). Show that the line segments AF and EC trisect the diagonal BD.
Sol.5) Given : A parallelogram ABCD, in which E and F are mid-points of sides AB and DC respectively.
To Prove : 𝐷𝑃 = 𝑃𝑄 = 𝑄𝐵
Proof : Since E and F are mid-points of AB and DC respectively.
⇒ 𝐴𝐸 = 12 𝐴𝐵 and 𝐶𝐹 = 12 𝐷𝐶        ...(i)
But, 𝐴𝐵 = 𝐷𝐶 and 𝐴𝐵 || 𝐷𝐶            ...(ii)       [Opposite sides of a parallelogram]
∴ 𝐴𝐸 = 𝐶𝐹 and 𝐴𝐸 || 𝐶𝐹.
⇒ 𝐴𝐸𝐶𝐹 is a parallelogram.              [One pair of opposite sides is parallel and equal]
In Δ𝐵𝐴𝑃, 𝐸 is the mid-point of 𝐴𝐵, 𝐸𝑄 || 𝐴𝑃
⇒ 𝑄 is mid-point of 𝑃𝐵                    [Converse of mid-point theorem]
⇒ 𝑃𝑄 = 𝑄𝐵                      ...(iii)
Similarly, in Δ𝐷𝑄𝐶,
P is the mid-point of DQ
𝐷𝑃 = 𝑃𝑄                        ...(iv)
From (iii) and (iv), we have
𝐷𝑃 = 𝑃𝑄 = 𝑄𝐵
or line segments 𝐴𝐹 and EC trisect the diagonal BD.                            Proved.

Q.6) Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Sol.6) Given : ABCD is a quadrilateral in which EG and FH are the line segments joining the midpoints of opposite sides.
To Prove : EG and FH bisect each other.
Construction : Join EF, FG, GH, HE and AC.
Proof : In ΔABC, E and F are mid-points of AB and BC respectively.
∴ 𝐸𝐹 = 1/2 𝐴𝐶 and 𝐸𝐹 || 𝐴𝐶                  ...(i)
In Δ𝐴𝐷𝐶, 𝐻 and G are mid-points of AD and CD respectively.
∴ 𝐻𝐺 = 1/2 𝐴𝐶 and 𝐻𝐺 || 𝐴𝐶                 ...(ii)
From (i) and (ii), we get
𝐸𝐹 = 𝐻𝐺 and 𝐸𝐹 || 𝐻𝐺
∴ 𝐸𝐹𝐺𝐻 is a parallelogram.                 [∵ a quadrilateral is a parallelogram if its one pair of
opposite sides is equal and parallel]
Now, EG and FH are diagonals of the parallelogram 𝐸𝐹𝐺𝐻.
∴ EG and FH bisect each other. [Diagonal of a parallelogram bisect each other]         Proved.

Q.7) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC.
(ii) 𝑀𝐷 ⊥ 𝐴𝐶
(iii) 𝐶𝑀 = 𝑀𝐴 = 1/2 𝐴𝐵
Sol.7) Given : A triangle ABC, in which ∠C = 90° and M is the mid-point of AB and 𝐵𝐶 || 𝐷𝑀.

To Prove : (i) D is the mid-point of AC                 [Given]
(ii) 𝐷𝑀 ⊥ 𝐵𝐶
(iii) 𝐶𝑀 = 𝑀𝐴 = 1/2 𝐴𝐵
Construction : Join CM.
Proof : (i) In Δ𝐴𝐵𝐶,
M is the mid-point of AB.         [Given]
𝐵𝐶 || 𝐷𝑀 [Given]
D is the mid-point of AC          [Converse of mid-point theorem] Proved.
(ii) ∠𝐴𝐷𝑀 = ∠𝐴𝐶𝐵                    [∵ Coresponding angles]
But ∠𝐴𝐶𝐵 = 90°                      [Given]
∴ ∠𝐴𝐷𝑀 = 90°
But ∠𝐴𝐷𝑀 + ∠𝐶𝐷𝑀 = 180°                 [Linear pair]
∴ ∠𝐶𝐷𝑀 = 90°
Hence, 𝑀𝐷 ⊥ 𝐴𝐶 Proved.
(iii) 𝐴𝐷 = 𝐷𝐶         ...(1)                 [∵ 𝐷 is the mid-point of AC]
Now, in Δ𝐴𝐷𝑀 and Δ𝐶𝑀𝐷, we have
∠𝐴𝐷𝑀 = ∠𝐶𝐷𝑀                 [Each = 90°]
𝐴𝐷 = 𝐷𝐶                  [From (1)]
𝐷𝑀 = 𝐷𝑀                 [Common]
∴ Δ𝐴𝐷𝑀 ≅ Δ𝐶𝑀𝐷 [SAS congruence]
⇒ 𝐶𝑀 = 𝑀𝐴                       ...(2)          [CPCT]
Since M is mid-point of AB,
∴ 𝑀𝐴 = 1/2 𝐴𝐵                  ...(3)
Hence, 𝐶𝑀 = 𝑀𝐴 = 1/2 𝐴𝐵 [From (2) and (3)]             Proved.