Q. 124.1( 430 Votes )

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

(ii) C = D

(iii) Δ ABC Δ BAD

(iv) Diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]


Answer :

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


AD||CE ( by construction)
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,


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)

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