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.]


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)


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