Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=\sqrt{(AE)^2-(AD)^2}$ also, $DE=AE\sin(EAD)$ and $BF=AF\sin(BAF)$ on the other hand we have $BAE+EAD=90^0$