TOPICS

Unit-9(Examples)

Quadrilaterals

**Example-1 :-** Show that each angle of a rectangle is a right angle.

Given that : Let ABCD be a rectangle in which ∠ A = 90°. Prove that : ∠ B = ∠ C = ∠ D = 90° Proof : We have, AD || BC and AB is a transversal. So, ∠ A + ∠ B = 180° (Interior angles on the same side of the transversal) But, ∠ A = 90° So, ∠ B = 180° – ∠ A = 180° – 90° = 90° Now, ∠ C = ∠ A and ∠ D = ∠ B (Opposite angles of the parallellogram) So, ∠ C = 90° and ∠ D = 90°. Therefore, each of the angles of a rectangle is a right angle.

**Example-2 :-** Show that the diagonals of a rhombus are perpendicular to each other.

Given that : ABCD is a Rhombus and AB = BC = CD = AD. Prove that : The diagonals of a rhombus are perpendicular to each other. Proof : Now, in Δ AOD and Δ COD, OA = OC (Diagonals of a parallelogram bisect each other) OD = OD (Common) AD = CD (Given) Therefore, Δ AOD ≅ Δ COD (SSS congruence rule) This gives, ∠ AOD = ∠ COD (CPCT) But, ∠ AOD + ∠ COD = 180° (Linear pair) So, 2∠ AOD = 180° or, ∠ AOD = 90° So, the diagonals of a rhombus are perpendicular to each other.

**Example-3 :-** : ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || AB. Show that

(i) ∠ DAC = ∠ BCA and (ii) ABCD is a parallelogram.

Given that : Δ ABC is isosceles. So, AB = AC. Prove that : (i) ∠ DAC = ∠ BCA and (ii) ABCD is a parallelogram. Proof : (i) Δ ABC is isosceles in which AB = AC (Given) So, ∠ ABC = ∠ ACB (Angles opposite to equal sides) Also, ∠ PAC = ∠ ABC + ∠ ACB (Exterior angle of a triangle) or, ∠ PAC = 2∠ ACB .....(1) Now, AD bisects ∠ PAC. So, ∠ PAC = 2∠ DAC .....(2) Therefore, 2∠ DAC = 2∠ ACB [From (1) and (2)] or, ∠ DAC = ∠ ACB (ii) Now, these equal angles form a pair of alternate angles when line segments BC and AD are intersected by a transversal AC. So, BC || AD Also, BA || CD (Given) Now, both pairs of opposite sides of quadrilateral ABCD are parallel. So, ABCD is a parallelogram.

**Example-4 :-** Two parallel lines l and m are intersected by a transversal p. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

Given that : Let ABCD be a rectangle in which ∠ A = 90°. Prove that : ∠ B = ∠ C = ∠ D = 90° Proof : We have, AD || BC and AB is a transversal. So, ∠ A + ∠ B = 180° (Interior angles on the same side of the transversal) But, ∠ A = 90° So, ∠ B = 180° – ∠ A = 180° – 90° = 90° Now, ∠ C = ∠ A and ∠ D = ∠ B (Opposite angles of the parallellogram) So, ∠ C = 90° and ∠ D = 90°. Therefore, each of the angles of a rectangle is a right angle.

**Example-5 :-** Show that the bisectors of angles of a parallelogram form a rectangle.

Given that : ABCD is a Rhombus and AB = BC = CD = AD. Prove that : The diagonals of a rhombus are perpendicular to each other. Proof : Now, in Δ AOD and Δ COD, OA = OC (Diagonals of a parallelogram bisect each other) OD = OD (Common) AD = CD (Given) Therefore, Δ AOD ≅ Δ COD (SSS congruence rule) This gives, ∠ AOD = ∠ COD (CPCT) But, ∠ AOD + ∠ COD = 180° (Linear pair) So, 2∠ AOD = 180° or, ∠ AOD = 90° So, the diagonals of a rhombus are perpendicular to each other.

**Example-6 :-** : ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and CD. If AQ intersects DP at S and BQ intersects CP at R, show that:

(i) APCQ is a parallelogram.

(ii) DPBQ is a parallelogram.

(iii) PSQR is a parallelogram.

Given that : ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and CD. Prove that : (i) APCQ is a parallelogram, (ii) DPBQ is a parallelogram and (iii) PSQR is a parallelogram. Proof : (i) In quadrilateral APCQ, AP || QC (Since AB || CD ).....(1) AP = 1/2 x AB, CQ = 1/2 x CD (Given) Also, AB = CD So, AP = QC ......(2) Therefore, APCQ is a parallelogram [From (1) and (2) and Theorem] (ii) Similarly, quadrilateral DPBQ is a parallelogram, because DQ || PB and DQ = PB. (iii) In quadrilateral PSQR, SP || QR (SP is a part of DP and QR is a part of QB) Similarly, SQ || PR So, PSQR is a parallelogram.

**Example-7 :-** In Δ ABC, D, E and F are respectively the mid-points of sides AB, BC and CA. Show that Δ ABC is divided into four congruent triangles by joining D, E and F.

Given that : D and E are mid-points of sides AB and BC of the Δ ABC Prove that : Δ ABC is divided into four congruent triangles by joining D, E and F. Proof : Given that D and E are mid-points of sides AB and BC of the Δ ABC, By mid point Theorem, DE || AC Similarly, DF || BC and EF || AB Therefore ADEF, BDFE and DFCE are all parallelograms. Now DE is a diagonal of the parallelogram BDFE, Therefore, Δ BDE ≅ Δ FED Similarly Δ DAF ≅ Δ FED and Δ EFC ≅ Δ FED So, all the four triangles are congruent.

**Example-8 :-** l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p. Show that l, m and n cut off equal intercepts DE and EF on q also.

Given that : AB = BC Prove that : DE = EF. Construction : Let us join A to F intersecting m at G.. Proof : The trapezium ACFD is divided into two triangles; i.e., Δ ACF and Δ AFD. In Δ ACF, it is given that B is the mid-point of AC (AB = BC) and BG || CF (since m || n). So, G is the mid-point of AF (by using mid point Theorem) Now, in Δ AFD, we can apply the same argument as G is the mid-point of AF, GE || AD and so By mid point Theorem, E is the mid-point of DF, i.e., DE = EF. In other words, l, m and n cut off equal intercepts on q also.

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