Let \[ABCD\] be a cyclic quadrilateral and \[PQRS\] be the quadrilateral formed by the angle bisectors of angle \[\angle A,\angle B,\angle C\text{ }and\angle D\]

Required to prove: \[PQRS\] is a cyclic quadrilateral.

Proof:

By angle sum property of a triangle

In \[\vartriangle APD\]

\[\angle PAD\text{ }+\angle ADP\text{ }+\angle APD\text{ }=\text{ }{{180}^{o}}~\ldots .\text{ }\left( i \right)\]

And, in \[\vartriangle BQC\]

\[\angle QBC\text{ }+\angle BCQ\text{ }+\angle BQC\text{ }=\text{ }{{180}^{o}}~\ldots .\text{ }\left( ii \right)\]

Adding (i) and (ii), we get

\[\angle PAD\text{ }+\angle ADP\text{ }+\angle APD\text{ }+\angle QBC\]

\[+\angle BCQ\text{ }+\angle BQC\text{ }=\text{ }{{180}^{o}}~+\text{ }{{180}^{o}}~=\text{ }{{360}^{o}}~\ldots \ldots \text{ }\left( iii \right)\]

But,

\[\angle PAD\text{ }+\angle ADP\text{ }+\angle QBC\text{ }+\angle BCQ\]

\[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }[\angle A\text{ }+\angle B\text{ }+\angle C\text{ }+\angle D]\]

\[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{360}^{o}}~=\text{ }{{180}^{o}}\]

Therefore,

\[\angle APD\text{ }+\angle BQC\text{ }=\text{ }{{360}^{o}}-\text{ }{{180}^{o}}~=\text{ }{{180}^{o}}\] [From (iii)]

But, these are the sum of opposite angles of quadrilateral \[PRQS\]

Therefore,

Quadrilateral \[PQRS\] is also a cyclic quadrilateral.