Question

If bisectors of opposite angles of a cyclic quadrilateral $ABCD$ intersect the circle, circumscribing it at the points $P$ and $Q$, prove that $PQ$ is a diameter of the circle.

Answer 1

Given:$ABCD$ is a cyclic quadrilateral.

$DP$ and $QB$ are the bisectors of $\angle D$ and $\angle B$, respectively.

To prove: $PQ$ is the diameter of a circle.

Proof: Since,$ABCD$ is a cyclic quadrilateral.

$\therefore \angle CDA+\angle CBA={180}^{\circ }$ since sum of opposite angles of cyclic quadrilateral is ${180}^{\circ }$

$\frac{1}{2}\angle CDA+\frac{1}{2}\angle CBA=\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$

$\Rightarrow \angle 1+\angle 2={90}^{\circ }$ ........$\left(1\right)$

But $\angle 2=\angle 3$ (angles in the same segment $QC$ are equal) ........$\left(2\right)$

$\Rightarrow \angle 1+\angle 3={90}^{\circ }$

From eqns$\left(1\right)$ and $\left(2\right)$,

$\angle PDQ={90}^{\circ }$

Hence,$PQ$ is a diameter of a circle, because diameter of the circle subtends a right angle at the circumference of the circle.