If the bisector of angle W and angle Y of a cyclic quadrilateral WXYZ meet at A and B respectively then prove that AB is the diameter of the circle. ?

Apr 28, 2018

Given

that the bisectors of angle W and angle Y of a cyclic quadrilateral WXYZ meet the circle at A and B respectively.
$A \mathmr{and} B$ are joined.

RTP

To prove that AB is the diameter of the circle.

Construction

$A \mathmr{and} Y$ are joined.

Proof

Sum of opposite angles of a cyclic quadrilateral being ${180}^{\circ}$ we have for cyclic quadrilateral $W X Y Z$

$\angle X W Z + \angle X Y Z = {180}^{\circ}$

$\implies \frac{1}{2} \angle X W Z + \frac{1}{2} \angle X Y Z = \frac{1}{2} \times {180}^{\circ}$

$\implies \angle X W A + \angle X Y B = {90}^{\circ}$ [ since WA and YB are bisectors of $\angle X W Z \mathmr{and} \angle X Y Z$ respectively.]

Bur $\angle X W A = \angle X Y A$, being the angles on same arc $A X$

So We have

$\angle X Y A + \angle X Y B = {90}^{\circ}$

$\implies \angle A Y B = {90}^{\circ}$

This means $A B$ must be diameter of the circle.