KöMaL 2013 C1201

An isosceles triangle AEB is drawn over side $AB$ of a square ABCD on the outside. The measure of its angle, at apex $E$, is $135^o$. Let the lines $AD$and $BE$ intersect at point $P$, and let $CE$ and $AB$ intersect at $Q$. Prove that $AP$=$BQ$.