Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ⊥ AB ⇒ ∠APO = ∠BPO = 90° But OX | | AB ∴ ∠POX = ∠APO [alternate interior angle] ⇒ ∠POX = 90° Similarly, ∠XOQ = 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved