Geometry and property of triangle

Question

$ABC$ is a cyclic triangle and bisector of angle $B\widehat{A}C$, $A\widehat{B}C$ and $A\widehat{C}B$ touches circle at $P$, $Q$ and $R$ respectively then measure of angle $R\widehat{Q}P$ is?

The options are:

1. $90-\frac{B\widehat{A}C}{2}$
2. $90-\frac{A\widehat{B}C}{2}$
3. $90+\frac{A\widehat{B}C}{2}$
4. $90+\frac{A\widehat{C}B}{2}$

I know how to draw the figure for this question but I can't establish the relation between the angles of the two triangles...

Answer 1

$\angle PQR=\angle PQB+\angle BQR=\angle BAP+\angle BCR=\frac A2+\frac C2=\frac{180-B}2=90-\frac B2$

Answer 2

The tool you need is an Inscribed angle theorem.

Roman83 shows how it applies to your problem.