Question

Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral ABCD is also cyclic.

$\left[4Marks\right]$

Answer 1

Given: A cyclic quadrilateral ABCD in which AP, BP, CR and DR are the bisectors of $\angle A,\angle B,\angle C$ and $\angle D$ respectively, forming a quadrilateral PQRS.

To prove: PQRS is a cyclic quadrilateral

Proof:

In $\Delta PAB$,

$\angle APB+\angle PAB+\angle PBA={180}^{\circ }$ [Angle sum property]

$\angle APB+\frac{1}{2}\angle A+\frac{1}{2}\angle B={180}^{\circ }$....(i)

$\angle PAB=\frac{1}{2}\angle Aand\angle PBA=\frac{1}{2}\angle B$
$\left[1Mark\right]$
In $\Delta RCD$,

$\angle CRD+\angle RCD+\angle RDC={180}^{\circ }$ [Angle sum property]

$\angle CRD+\frac{1}{2}\angle C+\frac{1}{2}\angle D={180}^{\circ }$....(ii)

$\angle RCD=\frac{1}{2}\angle Dand\angle RDC=\frac{1}{2}\angle D$
$\left[1Mark\right]$

$\angle APB+\angle CRD+\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)={360}^{\circ }$ Adding (i) and (ii)
$\left[1Mark\right]$

$\angle APB+\angle CRD+\frac{1}{2}\times {360}^{\circ }={360}^{\circ }$

$\angle APB+\angle CRD={180}^{\circ }$

Sum of a pair of opposite angles of quadrilateral PQRS is ${180}^{\circ }$

PQRS is a cyclic quadrilateral
$\left[1Mark\right]$
Hence proved.