Question

ABCD is a quadrilateral in which AB || DC and AD = BC (see figure). Show that:

(i) ∠A = ∠B and

(ii) ∠C = ∠D

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

Answer 1

Given: ABCD is a trapezium in which AB || CD and AD = BC.

To prove:

(i) ∠A = ∠B

(ii) ∠C = ∠D

Construction: Join AC and BD. Extend line AB and draw a line through C parallel to DA which meets AB produced at E.

Proof:

(i) AB || DC

⇒ AE || DC …(i)

and AD || CE …(ii) (by construction)

⇒ ADCE is a parallelogram

⇒ ∠A + ∠E = 180° …(iii) (sum of consecutive interior angles)

⇒ ∠ABC + ∠CBE = 180° …(iv) (linear pair of angles)

AD = CE …(v) (opposite sides of a parallelogram)

and AD = BC (given) …(vi)

⇒ BC = CE from (v) and (vi)

⇒ ∠CBE = ∠CEB …(vii) (angles opposite to equal sides)

⇒ ∠B + ∠E = 180° …(viii) [using (iv) and (vii)]

Now from (iii) and (viii), we have

∠A + ∠E = ∠B + ∠E

⇒ ∠A = ∠B

(ii) ∠A + ∠D = 180°

and ∠B + ∠C = 180°

⇒ ∠A + ∠D = ∠B + ∠C (∵ ∠A = ∠B)

⇒ ∠C = ∠D