# Geometry - angle bisector, circumcircle: SL olympiad

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 $\measuredangle A$. • WHY ARE YOU YELLING? Aug 18, 2015 at 9:26
• Why is this question on hold "as unclear"? It sounds pretty clear to me what is being asked. Aug 18, 2015 at 10:25
• I've added a picture to the question to make it clearer. If it's not the correct interpretation of the problem, please let me know what the correct interpretation is and I'll fix the picture. Aug 18, 2015 at 10:44
• So... what have you tried so far? Aug 18, 2015 at 10:54
• Sounds like a homework question, especially the Justify your answer from the original post. OP should add what he has tried already. Aug 18, 2015 at 12:13

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 $\measuredangle PRQ = 180^{\circ} - \measuredangle BPQ - \measuredangle CQP = \frac{1}{2}\measuredangle APQ + \frac{1}{2}\measuredangle AQP = 90^{\circ} - \frac{1}{2}\measuredangle A$.