Given : In trapezium ABCD, AB || CD and AD = BC.
To Prove :
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD
(iv) diagonal AC = diagonal BD
Constructions : Join AC and BD. Extend AB and draw a line through C parallel to DA meeting AB produced at E.
(i) Since AB || DC
⇒ AE || DC …(i)
and AD || CE …(ii) [Construction]
⇒ ADCE is a parallelogram [Opposite pairs of sides are parallel
∠A + ∠E = 180° …(iii) [Consecutive interior angles]
∠B + ∠CBE = 180° …(iv) [Linear pair]
AD = CE …(v) [Opposite sides of a ||gm]
AD = BC …(vi) [Given]
⇒ BC = CE [From (v) and (vi)]
⇒ ∠E = ∠CBE …(vii) [Angles opposite to equal sides]
∴ ∠B + ∠E = 180° …(viii) [From (iv) and (vii)
Now from (iii) and (viii) we have
∠A + ∠E = ∠B + ∠E
⇒ ∠A = ∠B Proved.
(ii) ∠A + ∠D = 180°
∠B + ∠C = 180°
⇒ ∠A + ∠D = ∠B + ∠C [∵ ∠A = ∠B]
⇒ ∠D = ∠C
Or ∠C = ∠D Proved.
(iii) In ∆ABC and ∆BAD, we have
AD = BC [Given]
∠A = ∠B [Proved]
AB = CD [Common]
∴ ∆ABC ≅ ∆BAD [ASA congruence]
(iv) diagonal AC = diagonal BD [CPCT] Proved
