Quadrilaterals with circles generated by internal bisectors

Question

Prove that intersections of internal bisectors of a quadrilateral form a cyclic quadrilateral.

Answer 1

Let's denote all angle with vertices. In triangle ADM we have:

$\frac{A}{2}+\frac{D}{2} + M=180$

In triangle BCP we have:

$\frac{B}{2}+\frac{C}{2}+ P=180$

Summing the two relation we get:

$\frac{A}{2}+\frac{D}{2} +\frac{B}{2}+\frac{C}{2} + M +P=360$

But due to definition we have:

$\frac{A}{2}+\frac{D}{2} +\frac{B}{2}+\frac{C}{2}=\frac{360}{2}= 180$

This means that:

$P+M=180$

That is P and M are supplementary which is specification of cyclic quadrilateral.