TITLE: Difficult triangle problem
QUESTION [8 upvotes]: In $△ABC$,  point $E$ on $AC$ such that $AE = 2 EC$ and $\angle ABE = 2 \angle EBC$, point $F$ on $BE$ such that $AF \perp BE$, and point $D$ on $AC$ such that $AD=DC$.
Prove that $DF \perp BC$. 
I (Michael) tried to prove that $\overrightarrow{DF}\cdot\overrightarrow{BC}=0$, but it gives very ugly computations.

REPLY [2 votes]: I have added a couple of points to the diagram:

Comparing areas says
$$
\overbrace{BC\cdot EB\sin(\delta)}^{\text{twice area of }BCE}=\overbrace{BA\cdot EB\sin(\delta)\cos(\delta)}^{\text{area of }ABE}
$$
Therefore,
$$
\cos(\delta)=\frac{BC}{BA}
$$
So if we let $BA=1$, we get $BC=\cos(\delta)$ and $BF=\cos(2\delta)$ and $AF=\sin(2\delta)$.
The Law of Cosines gives $AC=\sin(\delta)\sqrt{1+8\cos^2(\delta)}$, and then $AD=\frac12AC$, $DE=\frac16AC$, and $EC=\frac13AC$.
Similar triangles $\triangle AFE$ and $\triangle AGD$ give $GF=\frac14AF=\frac14\sin(2\delta)$.
The Pythagorean Theorem applied to $\triangle AEF$ gives $EF=\frac23\sin^2(\delta)$. Similar triangles give $DG=\frac12\sin^2(\delta)$.
Direct computation gives $\tan(\angle DFG)=\frac{DG}{GF}=\tan(\delta)$, so $\angle DFG=\delta$. This implies that $\angle HFB=\frac\pi2-\delta$, and therefore, $\angle FHB=\frac\pi2$. QED