# ABCD is a trapezi

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

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

However,

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

AB = BA (Common side)

B = A (Proved before)

ΔABC ΔBAD (SAS congruence rule)

ΔABC ΔBAD

AC = BD (By CPCT)

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