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