Argentine National Olympiad 2016

Write the condition $A{P}^2+3B{P}^2=A{Q}^2+3B{Q}^2$ in the form $A{Q}^2-A{P}^2=3(B{P}^2-B{Q}^2)$. 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

$B{P}^2-B{Q}^2=B{C}^2-B{D}^2-(C{P}^2-D{Q}^2)$. We showed above that $C{P}^2-D{Q}^2=A{Q}^2-A{P}^2$; on the other hand $A{Q}^2-A{P}^2=3(B{P}^2-B{Q}^2)$ by hypothesis. So the obtained equality can be written as $B{P}^2-B{Q}^2=B{C}^2-B{D}^2-3(B{P}^2-B{Q}^2)$, which implies $4(B{P}^2-B{Q}^2)=B{C}^2-B{D}^2$. Furthermore we have $\frac{BP}{BC}=\frac{BD}{BQ}=x$ with $x>0$ from the similar triangles $BCP$ and $BDQ$. Replacing $BC=xBP$ and $BD=xBQ$ in $4(B{P}^2-B{Q}^2)=B{C}^2-B{D}^2$ leads to $4(B{P}^2-B{Q}^2)=x^2(B{P}^2-B{Q}^2)$. Because $B{P}^2-B{Q}^2\ne 0$ and $x>0$, it follows that $x=2$. Then $BD=2BQ$, meaning that the hypotenuse $BD$ of right-angled triangle $BDQ$ is twice that its leg $BQ$. Therefore $\angle BDQ=30^{\circ }$ and so $A\hat{B}C=D\hat{B}Q=60^{\circ }$.