$\triangle ABC$ is a right-angled triangle at $A$. There are squares $AEDB$ and $ ACFG$ described from the outside on the legs $\overline{AB}$ and $\overline{AC}$, respectively. Line $CD$ meets the leg $\overline{AB}$ at $P$ and line $BF$ meets the leg $\overline{AC}$ at $Q$. Prove $|AP|=|AQ|$.
My attempt:
If we assume $|AP|=|AQ|$ indeed, then $\measuredangle APQ=\measuredangle PAD=45^\circ\implies\overline{PQ}\parallel\overline{FD}$.
Then $\triangle APQ$ and $\triangle PQF$ have a common edge $\overline{PQ}$ and equal altitudes and, therefore, $\operatorname{Area}(\triangle APQ)=\operatorname{Area}(\triangle PQF)$.
Let $S$ be the intersection point of $BF$ and $CD$. Then $$\boxed{\operatorname{Area}(\triangle DSQ)=\operatorname{Area}(\triangle PSF)}\tag 1$$
Since $BD\parallel AC$, $\triangle DBQ$ and $\triangle DBC$ with a common edge $\overline{BD}$ have equal altitude and, therefore, $\operatorname{Area}\triangle DBQ=\operatorname{Area}\triangle DBC$ $$\implies\operatorname{Area}(\triangle DSQ)=\operatorname{Area}(\triangle DBQ)-\operatorname{Area}(\triangle DBS)=\operatorname{Area}(\triangle DBC)-\operatorname{Area}(\triangle DBS)=\operatorname{Area}(\triangle BCS)$$ From $(1)$, it follows: $$\boxed{\operatorname{Area}(\triangle PSF)=\operatorname{Area}(\triangle BCS)}$$ Picture:
but I'm not sure if that information is relevant in this proof.
May I ask for advice on how to solve this task?
Thank you in advance!
Let $AB=c$ and $AC=b$.
Thus, $$\frac{AP}{DE}=\frac{AC}{EC},$$ which gives $$AP=\frac{bc}{b+c}.$$ By the same way $$AQ=\frac{bc}{b+c}$$ and we are done!