In right triangle $ABC,$ $AC = BC$ and $\angle C = 90^\circ.$ and Let $P$ and $Q$ be points on hypotenus $\overline{AB},$ as shown below, such that $\angle PCQ = 45^\circ.$ Show that \[AP^2 + BQ^2 = PQ^2.\]
\([asy] unitsize(3 cm); pair A, B, C, P, Q; A = (1,0); B = (0,1); C = (0,0); P = extension(C, C + dir(16), A, B); Q = extension(C, C + dir(16 + 45), A, B); draw(A--B--C--cycle); draw(C--P); draw(C--Q); label("$A$", A, SE); label("$B$", B, NW); label("$C$", C, SW); label("$P$", P, NE); label("$Q$", Q, NE); [/asy]\)
