A geometry problem: prove $AB=BF$.

Question

There is a question aksed by my friend, who is a teacher of a milde school. But I failed to give an answer to him. The problem is

In the right triangle $\Delta ABC$, the angle $\angle BAC=\frac{\pi}{2}$. The point $D$ is on $BC$ such that $AD\perp BC$. The point $E$ is the milde point of $CD$. Connecting $AE$, and let $F$ be the point on $AE$ such that $DF\perp AE$. Connecting $BF$, then proving that $AB=BF$.

I have no idea how to start, and I think it may be a very difficult problem. Looking forward to getting some hints, thanks!

Answer 1

Drop $BM\perp AE$ at $M$, and $BM$ intersects $AD$ at $H$, Connect $EH$. Since $H$ is the orthocenter of triangle $ABE$ we know $EH\perp AB$ and hence $EH \parallel AC$. Hence $DH = AH$. Hence $AM=MF$. Hence triangle $ABF$ is isosceles.

Answer 2

Well... brute force is always a solution? (The following derivation makes use of similar triangles heavily.)

Let $FG\perp BC$ for a point $G$ on $BC$. If we set $DF = a$, $AF = b$, and $AD = c$, i.e., $a^2+b^2 = c^2$, we would conclude that $DE = \frac{ac}{b}$. Therefore $CD = \frac{2ac}{b}$ and $BD = \frac{bc}{2a}$. Also note that $FG = \frac{a^2}{c}$, $DG = \frac{ab}{c}$, then we have

\begin{align} \overline{BF}^2 &= \overline{FG}^2 + \overline{BG}^2\\ &= \left(\frac{a^2}{c}\right)^2 + \left(\frac{bc}{2a} + \frac{ab}{c}\right)^2\\ &= \frac{a^4}{c^2} + \frac{b^2c^2}{4a^2} + \frac{bc}{a}\cdot \frac{ab}{c} + \frac{a^2b^2}{c^2}\\ &= \frac{a^4}{c^2} + \frac{b^2c^2}{4a^2} + b^2 + \frac{a^2b^2}{c^2}\\ &= \frac{a^4+a^2b^2}{c^2} + \frac{b^2c^2}{4a^2} + b^2\\ &= a^2 + \frac{b^2c^2}{4a^2} + b^2\\ &= c^2 + \frac{b^2c^2}{4a^2}\\ &=\left(\frac{bc}{2a}\right)^2 + c^2 \\ &= \overline{BD}^2 + \overline{AD}^2\\ &= \overline{AB}^2 \end{align}