Question

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that 

(i) ∠A = ∠B 

(ii) ∠C = ∠D 

(iii) ∆ABC ≅ ∠∆BAD 

(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer 1

Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. 

(i) AD = CE (Opposite sides of parallelogram AECD) 

However, AD = BC (Given) 

Therefore, BC = CE 

∠CEB = ∠CBE (Angle opposite to equal sides are also equal) 

Consider parallel lines AD and CE. AE is the transversal line for them. 

∠A + ∠CEB = 180º (Angles on the same side of transversal) 

∠A +∠CBE = 180º (Using the relation ∠CEB = ∠CBE) ... (1) 

However, ∠B + ∠CBE = 180º (Linear pair angles) ... (2) 

From equations (1) and (2), we obtain 

∠A = ∠B 

(ii) AB || CD 

∠A + ∠D = 180º (Angles on the same side of the transversal) 

Also, ∠C + ∠B = 180° (Angles on the same side of the transversal) 

∠A +∠ D =∠ C + ∠B 

However, ∠A = ∠B [Using the result obtained in (i)] 

∠C = ∠D 

(iii) In ∆ABC and ∆BAD, 

AB = BA (Common side) 

BC = AD (Given) 

∠B = ∠A (Proved before) 

∠∆ABC ∆∠BAD (SAS congruence rule) 

(iv) We had observed that, ∠∆ABC ∠∆BAD 

∠AC = BD (By CPCT)