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.