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A cyclic quadrilateral is given by the following figure:

A quadrilateral formed by the internal angle bisectors of a quadrilateral ABCD is PQRS:

We need to prove PQRS is cyclic.

A rule of angle equality is, vertically opposite angles are equal.

Angle sum property of a triangle gives the sum of all angles of a triangle = ${180^0}$

To prove the quadrilateral PQRS is cyclic, it is enough to prove that the sum of opposite angles of PQRS is ${180^0}$ . Implies $\angle SPQ + \angle SRQ = {180^0}$

Step 1: Consider a quadrilateral ABCD with internal bisectors AQ, BS, CS, DQ of angles $\angle A$, $\angle B$ , $\angle C$ and $\angle D$ respectively. We need to prove the quadrilateral PQRS formed by these four internal angle bisectors is cyclic.

As vertically opposite angles are equal, we obtain the following equalities:

$

\angle SPQ = \angle APB \\

\angle SRQ = \angle DRC \\

$

Adding above two formulas we get $\angle SPQ + \angle SRQ = \angle APB + \angle DRC$

As AQ is the angle bisector of $\angle A$ and P is a point on AQ, By angle sum property,

$\angle APB = 180 - (\dfrac{1}{2}\angle A + \dfrac{1}{2}\angle B)$ and $\angle DRC = 180 - (\dfrac{1}{2}\angle D + \dfrac{1}{2}\angle C)$ . Thus,

$

\angle SPQ + \angle SRQ = \angle APB + \angle DRC \\

= 180 - (\dfrac{1}{2}\angle A + \dfrac{1}{2}\angle B) + 180 - (\dfrac{1}{2}\angle D + \dfrac{1}{2}\angle C) \\

= 360 - \dfrac{1}{2}(\angle A + \angle B + \angle C + \angle D) \\

= 360 - \dfrac{1}{2}(360) \\

= {180^0} \\

$

The sum of opposite angles of quadrilateral PQRS is ${180^0}$ . Thus PQRS is a cyclic quadrilateral.

Hence proved.

The quadrilateral formed by internal angle bisectors of a quadrilateral is cyclic.

Kite, Trapezoid, Parallelogram, Square, Rhombus, Rectangle comes under Quadrilateral.