Difficult triangle problem 
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.
 A: (I've renamed a couple of points. $P$ is the foot of the perpendicular from $A$ to $\overline{BE}$, and $M$ is the midpoint of $\overline{AC}$. I've also conveniently subdivided $\overline{AC}$ into sixths, to clearly indicate both trisection by $E$, bisection by $M$, and the key ratio in $(3)$.)

Drop a perpendicular from $C$ to $Q$ on $\overleftrightarrow{BE}$. Then, $$\triangle APE \sim \triangle CQE \quad\text{with}\quad |\overline{AE}|:|\overline{CE}| = 2:1 \tag{1}$$ 
so that
$$\frac{2}{1} \;=\; \frac{|\overline{AP}|}{|\overline{CQ}|} \;=\; \frac{|\overline{AB}| \sin 2\delta}{|\overline{BC}|\sin\delta} \;=\; \frac{|\overline{AB}|\cdot 2 \sin\delta\cos\delta}{|\overline{BC}|\cdot\sin\delta} \quad\to\quad |\overline{BC}| = |\overline{AB}|\cos\delta \tag{2}$$
As @Alfred did, we'll extend $\overline{BE}$ to a point (we'll call ours $A^\prime$) such that $\overline{AB}\cong\overline{A^\prime B}$. By $(1)$, we know that $\triangle A^\prime B C$ has a right angle at $C$; moreover, this triangle is congruent to both $\triangle ABC^\prime$ and $\triangle A^\prime B C^\prime$, where $C^\prime$ is the midpoint of $\overline{AA^\prime}$.

Note that $\overline{C^\prime C} \parallel \overline{AP}$ (as both are perpendicular to $\overline{PA^\prime}$), and $\overline{C^\prime C}$ meets the midpoint of $\overline{PA^\prime}$ (by, say, the Midsegment Theorem applied to $\triangle APA^\prime$). We may conclude that $\square PCA^\prime C^\prime$ is a rhombus, which implies $\overline{AC^\prime} \cong \overline{CP}$. Consequently, $\square APCC^\prime$ is a parallelogram, with mutually-bisecting diagonals: $M$, the midpoint of $\overline{AC}$, coincides with the midpoint of $\overline{PC^\prime}$. Thus, $P$, $C^\prime$, $M$ are collinear, with a shared line that, being parallel to $\overline{A^\prime C}$, must be perpendicular to $\overline{BC}$. $\square$
A: I can give an elegant non brute force solution.
Lets say: 
$$\operatorname{area}(CBE)=0.5\times CB\times BE\times\sin \delta$$ 
and 
$$\operatorname{area}(ABE)=0.5\times AB\times BE\times \sin(2\delta).$$ 
Then the ratio of areas is
$$\frac{CB}{2AB\cos(\delta)}=CE/AE=1/2.$$ 
Thus we conclude $\cos\delta=\frac{CB}{AB}$. We can then extend $BE$ past to a point $X$ so that $BX=AB$. In the isosceles triangle $ABX$ we let Y be the midpoint of $AX$. If $DF$ intersects $CB$ at $Q$ we need only to prove that $QBY$ is similar to $AFB$.
This is easy to see (after 3 hours of work) because if $BY$ intersects $AF$ at $T$, then $ATB$ is similar to $YFB$ because they share $\delta$ as angle and $FB/BT=BY/AB=\cos\delta$. 
A: A possible approach based on analytic geometry and some trigonometry. Let us set a Cartesian plane with the origin in $E$ and the $y$-axis aligned with the $BE$ segment. Let us call $(0,Y_B)$ the coordinates of point $B$.  The equations of the lines $BC$ and $BA$ are then $$y=\tan(\pi/2-\beta)x+Y_B$$ $$y=\tan(2 \beta-\pi/2)x+Y_B$$ respectively. 
Now let us call  $y=mx$ the equation of the line corresponding to the segment $CA$. Thus, the coordinates of $C$ are 
$$\left(\frac{Y_B}{m-\tan{\beta}}, \frac{mY_B}{m-\tan{\beta}}\right)$$
and those of $A$ are
$$\left(\frac{Y_B}{m-\tan{-2 \beta}}, \frac{mY_B}{m-\tan{-2 \beta}}\right)$$
Now calculating the length of $CE$ and$ EA$ by the standard formulas, and substituting in $CE=1/2 \,EA \,\,$ we get 
$$ \left[(\frac{Y_B}{m-\tan(\pi/2- \beta)})^2 + ( \frac{m Y_B}{m-\tan( \pi/2-\beta)})^2\right]=    \frac{1}{4}        \left[(\frac{Y_B}{m-\tan(2 \beta-\pi/2)})^2 + ( \frac{m Y_B}{m-\tan(2 \beta-\pi/2)})^2\right]     $$
Expanding and developing the calculations, this equation is considerably simplified in 
$$-[m - \cot(\beta)]= 2 [\cot(2 \beta) + m]$$
and then
$$m=\frac{\tan{\beta}}{3}$$
which is the slope of $CA$. Since now we know the equations of both $CA$ and $BA$, we can calculate the coordinates of their intersection $A$, which are found to be
$$\left(\frac{3 Y_B}{3 \cot(2 β) + \tan(\beta)} , \frac{\tan(\beta) \,Y_B}{3 \cot(2 β) + \tan(\beta)}\right) $$
Knowing this, it remains to be calculated the slope of $DF$. Because $ED= \frac{1}{4} EA \,\,$, the coordinates of the point $D$ are $1/4$ of those of the point $A$ shown above. On the other hand, the point $F$ is on the $y$-axis, and because $AF$ is perpendicular to $EF$,  its $y$-coordinate is the same of point $A$. From this, we get that the slope of $DF$ is 
$$-\frac{3/4 \,(\tan(\beta) Y_B)/(3 \cot(2 β) + \tan(\beta))}{    
1/4 \, (3 Y_B)/(3 \cot(2 β) + \tan(\beta)}$$
$$=-\tan{\beta}$$
This slope corresponds to a line that is perpendicular to the segment $BC$, since 
$$\frac{1}{\tan(\beta)}=\tan(\pi/2-\beta)$$
A: I will give you the "brute force" approach, and leave you to do the computation if you should choose this approach. I have a slicker approach i might try later on, but for now i must leave. The idea is to use coordinate geometry because inner products are an easy characterization of orthogonality:
Take $A = (0,0) \in \mathbb{R}^2$, and wlog (since the problem is invariant under scaling), take $C = (1,0)$, so that $E = (\frac{2}{3},0)$. Write $B = (x,y)$.
The line $BE$ is then $L(t) = (\frac{2}{3}, 0) + t (x-\frac{2}{3}, y)$.
$F = L(s)$ for some $s$ which we will find. At $F$, $<F-A, B-E> = 0$. Since $A = 0$, you get an equation for $s$ which is $<L(s), B-E> = 0$. Solve this equation. Now you know the point $F$.
Note now that $D = (\frac{1}{2},0)$. So you know $F - D$. $B - C = (x-1,y)$ Compute $<F - D, B-C>$. You should get $0$, which would complete the proof.
A: Since $AD=DC$ and $EC=2AE$ we get that 
$$(A,E;D,C)=\frac{\vec{AD}}{\vec{DE}}:\frac{\vec{AC}}{\vec{CE}} = 2 : (-2) = -1,$$
so $A,E,D,C$ is a harmonic quadruple. Along with $\angle EFA=90^\circ$ this implies that $FE$ is the bisector of angle $CFD$. Angle bisector theorem yields
$$\frac{FC}{FD} = \frac{EC}{ED} = 2.$$
Let $G$ be the point symmetric to $A$ with respect to the line $BE$. Then $F$ is the midpoint of $AG$ and since $D$ is the midpoint of $AC$ we have $CG = 2DF = FC$. 
Also note that 
$$\angle GBC = \angle GBF - \angle CBF = \angle EBA - \delta = 2\delta - \delta = \delta = \angle CBE,$$
so $BC$ is the bisector of the angle $GBF$.
It follows that $C$ is the intersection of the angle bisector of $GBF$ and the perpendicular bisector of $FG$. A well-known lemma states that $C$ lies on the circumcircle of $BFG$. In particular $\angle GCB = \angle GFB = 90^\circ$.
Finally, since $CG \parallel DF$ and $CG \perp BC$, we have $DF \perp BC$. 
A: 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
