Given: ABCD is a cyclic quadrilateral. The angle bisectors of ∠DAB and ∠BCD intersect the circle at points X and Y respectively.
To prove: XY is the diameter of the circle.
Construction: Join points A and Y.
Proof: ∠YAB and ∠YCB are angles subtended by the same arc YB of the circle. ∴ ∠YAB = ∠YCB = ½∠BCD -----(i) (since CY is the bisector of ∠BCD)
Now, ∠XAY = ∠XAB + ∠YAB = ½(∠BAD) + ½(∠BCD) (from (i) and since AX is the bisector of ∠DAB) = ½(∠BAD + ∠BCD) = ½ × 180° (since opposite angles of a cyclic quadrilateral sum to 180°) = 90°
∴ ∠XAY is a right angle subtended in a semicircle. ∴ XY is a diameter of the circle.