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 = 180^{0} (Angles on the same side of transversal)

∠A + ∠CBE = 180^{0} (Using the relation CEB = CBE) ...(1)

However,

∠B + ∠CBE = 180^{0} (Linear pair angles) ...(2)

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

∠A = ∠B

**(ii)** AB || CD

∠A + ∠D = 180^{0} (Angles on the same side of the transversal)

Also,

∠C + ∠B = 180^{0} (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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