ABCD is a cyclic quarilateral. The bisectors of `angleaandangleC` in tersect the circle at the points E and F respectively . Prove that EF is a diameter of the circle.

Explanation

Let ABCD is a cyclic quadrilateral in the circle with centre at O . The bisector AE of the `angleA` intersects the circle at E and the bisector of the `angleC` intersects the circle at F .Let us join E F
To Prove : EF is a diameter of the circle.
Construction : Let us join C, E.
Proof : ABCD is a cyclic quadrilateral.
`thereforeangleBAD+angleBCD=180^(@)`
or. `(1)/(2)angleBAD+(1)/(2)angleBCD=90^(@)` [dividing by 2 ]
or, `angleDAE+angleDCF=90^(@) " " [because` BE and CF are the bisectors of `angleAandangleC` respectively]
But two angles in circle produced by the arc DE are `angleDAEand angleDCE`.
`therefore" " angleDAE=angleDCE`
`therefore` from (1) we get , `angleDCE+angleDCF=90^(@)" or " angleECF=90^(@)`
`therefore" " angleECF` is semicircular angle.
Hence EF is a diameter of the circle .