Q.ABCD is a cyclic quadrilateral. The angle bisectors of \(\angle\)DAB and \(\angle\)BCD intersect the circle at points X and Y respectively. If O is the center of the circle, find the value of \(\angle\)XOY.

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.
Similar Questions