ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that

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 = 180⁰ (Angles on the same side of transversal)

∠A + ∠CBE = 180⁰ (Using the relation CEB = CBE) (1)

However,

∠B + ∠CBE = 180⁰ (Linear pair angles) (2)

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

∠A = ∠B

(ii) AB || CD

∠A + ∠D = 180⁰ (Angles on the same side of the transversal)

Also,

∠C + ∠B = 180⁰ (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 ΔABC and ΔBAD,

AB = BA (Common side)

BC = AD (Given)

∠B = ∠A (Proved before)

ΔABC ≅ ΔBAD (SAS congruence rule)

(iv) We had observed that,

ΔABC ≅ ΔBAD

AC = BD (By CPCT)