If bisectors of opposite angles of a cyclic

Solution:

Given, ABCD is a cyclic quadrilateral.

DP and QB are the bisectors of ∠D and ∠B, respectively.

To prove PQ is the diameter of a circle.

Construction Join QD and QC.

Proof Since, $A B C D$ is a cyclic quadrilateral.

$\therefore \quad \angle C D A+\angle C B A=180^{\circ}$

[sum of opposite angles of cyclic quadrilateral is $180^{\circ}$ ]

On dividing both sides by 2 , we get

$\frac{1}{2} \angle C D A+\frac{1}{2} \angle C B A=\frac{1}{2} \times 180^{\circ}=90^{\circ}$

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

$\left[\angle 1=\frac{1}{2} \angle C D A\right.$ and $\left.\angle 2=\frac{1}{2} \angle C B A\right]$

But $\angle 2=\angle 3$ [angles in the same segment $Q C$ are equal] ... (ii)

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

From Eqs. (i) and (ii), $\angle P D Q=90^{\circ}$

Hence, $P Q$ is a diameter of a circle, because diameter of the circle. Subtends a right angle at the circumference.