Difficult triangle problem
(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 $\overline{BC}$. $\square$
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 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
$$\frac{1}{\tan(\beta)}=\tan(\pi/2-\beta)$$