International Mathematical Olympiad

If $AC=BC$, $△ABC$ is an isosceles triangle, and $CR$ is the symmetry axis of $△RQL$ and $△RPK$. The conclusion is obviously true.

If $AC\ne BC$, without loss of generality, let $AC<BC$. Denote the center of circumcircle of $△ABC$ by $O$.

Since the right triangles $CQL$ and $CPK$ are similar, $$\angle CPK=\angle CQL=\angle OQP,\text{ and }\frac{QL}{PK}=\frac{CQ}{CP}.①$$ Let $l$ be the perpendicular bisector of $CR$, then $O$ is on $l$. Since $△OPQ$ is an isosceles triangle, $P$ and $Q$ are two points symmetrical about $l$ on $CR$. So $$RP=CQ\text{and}RQ=CP.\stackrel{◯}{2}$$ By ①, ②, $$\begin{array}{rl}\frac{S(△RQL)}{S(△RPK)}& =\frac{\frac{1}{2}\cdot RQ\cdot QL\cdot \sin \angle RQL}{\frac{1}{2}\cdot RP\cdot PK\cdot \sin \angle RPK}\\ & =\frac{RQ}{RP}\cdot \frac{QL}{PK}=\frac{CP}{CQ}\cdot \frac{CQ}{CP}=1.\end{array}$$ Hence the two triangles have the same area.