Geometry - angle bisector, circumcircle: SL olympiad

Question

I tried this problem as much as I can, but I got nothing. This is a Sri Lankan mathematical olympiad problem.

Let $P$,$Q$ be points on the sides $AB$ and $AC$, respectively, of a $\triangle ABC$ ,which satisfy $BP+CQ = PQ$. Let $R$ be the point of intersection, other than $A$, of the bisector of the $\measuredangle BAC$ and the circumcircle of the $\triangle ABC$. Express $\angle PRQ$ in terms of $\measuredangle A$.

Answer 1

The main observation is that R is the center of an excircle of $\triangle APQ$. Also note that because $\measuredangle BAR = \measuredangle CAR$ we have BR=CR s.t. R can also be defined as the intersection of the bisector of $\angle A$ and the bisector of BC.

Let R' be the intersection of the angle bisectors of $\angle QPB$ and $\angle PQC$. Since R' is the center of an excircle of $\triangle APQ$, R' is also on the bisector of $\angle PAQ$.

Let S be on PQ s.t. PS = PB and QS = QC (this is possible because PQ = PB + QC). Now $\triangle PR'S$ is congruent to $\triangle PR'B$ (SAS), so R'B = R'S. Analogously R'C = R'S. So R'B = R'C.

Now we have R' on the bisector of $\angle PAQ$ and R' on the bisector of BC, so R'=R, i.e. R is the center of an excircle of $\triangle APQ$. Hence $\measuredangle PRQ = 180^{\circ} - \measuredangle BPQ - \measuredangle CQP = \frac{1}{2}\measuredangle APQ + \frac{1}{2}\measuredangle AQP = 90^{\circ} - \frac{1}{2}\measuredangle A$.