In right triangle \(ABC\), \(AC = BC\) and \(\angle C = 90^\circ\). Let \(P\) and \(Q\) be points on hypotenuse \(\overline{AB}\), as shown below, such that \(\angle PCQ = 45^\circ\). Show that \(AP^2 + BQ^2 = PQ^2.\)
Hi! Your answer is very clear, but I'm not entirely sure this works since it looks like your solution relies on ∠BCQ ≅ ∠ACP, which may not always be the case.
