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.
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.