Question

Question 9
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the point P and Q, prove that PQ is a diameter of the circle.

Answer 1

Construction: Join QD and QC.

ABCD is a cyclic quadrilateral.
$\therefore \angle CDA+\angle CBA={180}^{\circ }$ [sum of opposite angles of cyclic quaddrilaterl is ${180}^{\circ }$]
On dividing both sides by 2, we get
$\frac{1}{2}\angle CDA+\frac{1}{2}\angle CBA=\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$
$\Rightarrow \angle 1+\angle 2={90}^{\circ }$ ...(i)
$[\angle 1=\frac{1}{2}\angle CDAand\angle 2=\frac{1}{2}\angle CBA]$
But $\angle 2=\angle 3$ [angles in the same segment QC are equal] ….(ii)
From Eqs. (i) and (ii), $\angle PDQ={90}^{\circ }$
Hence, PQ is a diameter of a circle, because diameter of the circle subtends a right angle at the circumference.