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

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

Since the right triangles $CQL$ and $CPK$ are similar,

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