Given: ABCD is a trapezium, in which AB || DC and AD = BC. To Prove: (i) ∠A = ∠B (ii) ∠C = ∠D (iii) △ABC ≅ △BAD (iv) Diagonal AC = diagonal BD. Const.: Produce AB to E, such that a line through C parallel to DA (CE || DA), intersects AB in E. Proof : (i) In quad. AECD AE || DC [given] and AD || EC [by const.] ⇒ AECD is a Parallelogram ⇒ AD = EC [opp side ||gm] But, AD = BC [given] ⇒ BC = EC ⇒ ∠2 = ∠1 [angle opp. to equal sides of a △] Also, ∠1 + ∠3 = 180° [linear pair] and ∠A + ∠2 = 180° [consecutive interior ∠s] ⇒ ∠A + ∠2 = ∠1 + ∠3 ⇒ ∠A = ∠3 [∵ ∠2 = ∠1] or ∠A = ∠B (ii) AB || DC and AD is a transversal ∴ ∠A + ∠D = 180° [consecutive interior angle] ⇒ ∠B + ∠D = 180° ---(i)[∵∠A = ∠B] Again, AB || DC and BC is transversal . ∴ ∠B + ∠C = 180° From (i) and (ii), we have ∠B + ∠C = ∠B + ∠D ⇒ ∠C = ∠D (iii) In△ABC and △BAD we have, AB = AB [common] BC = AD [given] ∠ABC = ∠BAD [proved above] ⇒ △ABC ≅ △BAD [by SAS congruence axiom] (iv) ⇒ AC = BD [c.p.c.t.] Thus, diagonal AC = diagonal BD.