Argentine National Olympiad

$$A{P}^2+3B{P}^2=A{Q}^2+3B{Q}^2$$ $$A{Q}^2-A{P}^2=3(B{P}^2-B{Q}^2)$$ Express $A{Q}^2$ and $A{P}^2$ by Pythagoras theorem for the right-angled triangles $ADQ$ and $ACP$: $A{Q}^2=A{D}^2-D{Q}^2$, $A{P}^2=A{C}^2-C{P}^2$. Since $AC=AD$, it follows that $A{Q}^2-A{P}^2=C{P}^2-D{Q}^2$. Likewise the right-angled triangles $BCP$ and $BDQ$ yield $B{P}^2=B{C}^2-C{P}^2$, $B{Q}^2=B{D}^2-D{Q}^2$. Hence