ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: (i) ∠A = ∠B (ii) ∠C = ∠D (iii) △ABC ≅ △BAD (iv)…
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.