If two sides of a cyclic-quadrilateral are parallel; prove that :

(i) its other two sides are equal.

(ii) its diagonals are equal.

Let ABCD is a cyclic quadrilateral in which AB || DC. AC and BD are its diagonals.

As AB || DC (given)

∠DCA = ∠CAB [Alternate angles are equal]

Chord AD subtends ∠DCA and chord BC subtends ∠CAB at the circumference of the circle.

and

∠DCA = ∠CAB

We know that,

If the angles subtended by 2 chords on the circumference of the circle are equal, then the lengths of the chords are also equal.

∴ chord AD = chord BC or AD = BC.

Hence, proved that AD = BC.

(ii) From figure,

⇒ ∠A + ∠C = 180° [As, sum of opposite angles in a cyclic quadrilateral = 180°]

Also,

⇒ ∠B + ∠C = 180° [Sum of co-interior angles = 180° (As, AB || CD)]

∴ ∠B + ∠C = ∠A + ∠C

⇒ ∠B = ∠A

In ∆ABC and ∆ADB

⇒ AB = AB [Common]

⇒ ∠B = ∠A [Proved above]

⇒ BC = AD [Proved above]

Hence, by SAS criterion of congruence

∆ACB ≅ ∆ADB

∴ AC = BD [By C.P.C.T.]

Hence, proved that AC = BC.