Bisectors of angle A & C of a cyclic quadrilateral ABCD intersect a circle through A,B,C,D at E&F respectively. Proove that EF is the diameter of the circle.

Given: ABCD is a cyclic quadrilateral. AE and CF are the bisectors of ∠A and ∠C respectively.

To prove: EF is the diameter of the circle i.e. ∠EAF = 90°

Construction: Join AF and FD.

Proof:

ABCD is a cyclic quadrilateral.

∴ ∠A + ∠C = 180°         (Sum of opposite angles of a cyclic quadrilateral is 180° )

⇒ ∠EAD + ∠DCF = 90°     ...(1)          (AE and CF are the bisector of ∠A and ∠C respectively)

∠DCF = ∠DAF      ...(2)       (Angles in the same segment are equal)

From (1) and (2), we have

∠EAD + ∠DAF = 90°

⇒ ∠EAF = 90°

⇒ ∠EAF is the angle in a semi-circle.

⇒ EF is the diameter of the circle.