Given an equilateral triangle $ABC$ and $P$ is any point inside the triangle such that $PA^2 = PB^2 + PC^2$. Then what is angle $BPC$?

Question

I could only come up with a rough diagram but I couldn't move any further.

Answer 1

Let $Q$ be the image of $P$ by the $60$-degree rotation that turns $A$ to $B$. Then $\triangle ACP = \triangle BCQ$, and we have $BQ = AP$. That, together with $PQ = CP$, implies that $PB^2 + PQ^2 = BQ^2$. Now you can apply the inverse Pythagorean theorem to see $\angle BPQ = 90^\circ$. And don't forget that $\angle QPC$ is an angle of an equilateral triangle.

Answer 2

This is the solution of the problem. I think that it is self explaining, but if you have any question feel free to ask

hope this helps