**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^{o}

**Construction:** Join AE and FD.

**Proof:**

ABCD is a cyclic quadrilateral.

∴ ∠A + ∠C = 180^{o} (Sum of opposite angles of a cyclic quadrilateral is 180^{o} )

⇒ ∠EAD + ∠DCF = 90^{o} ...(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^{o}

⇒ ∠EAF = 90^{o}

⇒ ∠EAF is the angle in a semi-circle.

⇒ EF is the diameter of the circle.