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Another nice solution at https://stanfulger.blogspot.com/2021/12/tuymaada-junior-2006.html Best regards, sunken rock

Let $D$ be on $\overline{AC}$ and let $E$ be on $\overline{BC}$ with $\overline{BC}$ being a horizontal line parallel to the x-axis. We proceed with coordinate bashing. Let point $C$ be the origin. We then let $B=(a,0)$, $A=(0,a)$, $E=(b,0)$, and $D=(0,b)$. We then proceed by finding the equations of a few lines. $$\overline{AB} - y=-x+a$$$$\overline{AE} - y=\frac{-a}{b}x+a$$Since $\overline{CP}$ and $\overline{DQ}$ are perpendicular to $\overline{AE}$, they both have a slope of $\frac{b}{a}$. We then use points $D$ and $C$ to find the equations of these two lines. $$\overline{PC} - y=\frac{b}{a}x$$$$\overline{DQ} - y=\frac{b}{a}x+b$$Next, we find that the point of intersection of $\overline{AB}$ and $\overline{PC}$ (which is point $P$) is $(\frac{a^2}{a+b}, \frac{ab}{a+b})$. We then find that the intersection of $\overline{AB}$ and $\overline{DQ}$ (which is point $Q$) is $(\frac{a^2-ab}{a+b}, \frac{2ab}{a+b})$. The final thing to do is find the distances $BP$ and $PQ$ (we will multiply the lengths by $(a+b)$ for simplicity). $$(a+b)(BP)=\sqrt{(a^2+ab-a^2)^2+(0-ab)^2}=\sqrt{2a^2b^2}$$$$(a+b)(QP)=\sqrt{(a^2-a^2+ab)^2+(ab-2ab)^2}=\sqrt{2a^2b^2}$$Since $BP=QP$, our proof is complete. $\quad \square$