Iscoceles problem

Let $C$ be the apex of isosceles triangle $\triangle ABC$ and let $P$ be on $\overline{AB}$ with $R$ and $Y$ as defined in the problem (see the diagram below).

Let $x=AC=BC$ be the length of the legs of the triangle. Note that the segment $\overline{PC}$ divides $\triangle ABC$ into two triangles. Thus $\begin{aligned} \mathrm{Area}(\triangle ABC) &= \mathrm{Area}(\triangle APC) + \mathrm{Area}(\triangle BPC) \\ &= AC\times PR + BC\times PY \\ &= x(PR+PY) \end{aligned}$

Therefore $PR+PY = \dfrac{\mathrm{Area}(\triangle ABC)}{x}$ . Since both the area of $\triangle ABC$ and $x$ (the length of the legs) are constant for a given triangle, $PR+PY$ is constant as well. The statement is True .