In the square $ABCD$, prove that $BF+DE=AE$.

Question

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 $BAE+EAD=90^0$

Answer 1

Let $AB=x$ and angle $<BAF=<FAE=\alpha=> DE=x\tan(\frac{\pi}{2}-2\alpha),\; BF=x\tan\alpha=>$ $$ \\BF+DE=x(\tan(\frac{\pi}{2}-2\alpha)+\tan\alpha)=x(\cot2\alpha+\tan\alpha) \\AE=\frac{x}{\cos(\frac{\pi}{2}-2\alpha)}=\frac{x}{\sin2\alpha} \\BF+DE=AE<=>\cot2\alpha+\tan\alpha=\frac{1}{\sin2\alpha}<=> \\\cos2\alpha=1-2\sin^2\alpha $$

Answer 2

Hint: Rotate the $F$ around $A$ for $90^{\circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)