The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 6

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.