**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)…**

###### Answers:

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.