Question

Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.

Answer 1

To prove: PQRS is cyclic

$AP,BP,CR,DR$ are the angle bisector of $\angle A,\angle B,\angle C,\angle D.$ respectively.

We know that the sum of all three angles of a triangle is ${180}^{\circ }$.

$\angle SPQ={180}^{\circ }-\left(\frac{a+b}{2}\right)$ [from $\Delta PAB$]

$\angle SRQ={180}^{\circ }-\left(\frac{d+c}{2}\right)$ [from $\Delta DRC$]

$\therefore \angle SPQ+\angle SRQ={360}^{\circ }-\left(\frac{a+b+c+d}{2}\right)$

But $a+b+c+d={360}^{\circ }$ (Sum of internal angles of quadrilateral)

$\therefore \angle SPQ+\angle SRQ={360}^{\circ }-\frac{{360}^{\circ }}{2}={180}^{\circ }$

Similarly $\angle PQR+\angle PSR={180}^{\circ }$

These angles are opposite angles of quadrilateral PQRS

$\therefore$ PQRS is cyclic. [When the sum of opposite angles of quadrilaterals is ${180}^{\circ },$ then that is a cyclic quadrilateral ]