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

(i) ∠ A = ∠ B

(ii) ∠ C = ∠ D

(iii) ∆ ABC ≅ ∆ BAD

(iv) diagonal AC = diagonal BD

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

Solution:

To Construct: Draw a line through C parallel to DA intersecting AB produced at E.

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

AD = BC (Given)

BC = CE

⇒∠CBE = ∠CEB

also, ∠A+∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)

∠B +∠CBE = 180° ( As Linear pair)

⇒∠A = ∠B

(ii) ∠A+∠D = ∠B+∠C = 180° (Angles on the same side of transversal)

⇒∠A+∠D = ∠A+∠C (∠A = ∠B)

⇒∠D = ∠C

(iii) In ΔABC and ΔBAD

AB = AB (Common)

∠DBA = ∠CBA

AD = BC (Given)

, ΔABC ≅ ΔBAD [SAS congruency]

(iv) Diagonal AC = diagonal BD [by CPCT] as  ΔABC ≅ ΔBAD.