In the following figure $AF=BD=DC$ and $AE=EF$. Find the angle $\alpha$.
I tried so many different things (constructing triangles, drawing parallel lines, using Ceva theorem in triangles $ABD$ and $CBE$, ...). Any help for solving this will be much appreciated.
A geomtrical solution (without trigonometry) would be very nice.
In isosceles $\triangle AEF$, drop $EG \perp AF$ with $G$ on $AF$. Let $AG=GF=x$. So $BD=2x$. Let $\angle BAD = \theta$. So $AE=x\sec \theta$.
Applying Menelaus' for transversal $CE$ to $\triangle ABD$, $$\frac{BC}{DC}\cdot\frac{DF}{FA}\cdot\frac{AE}{EB}=1$$ $$\Rightarrow DF=EB\cos \theta$$ $$\Rightarrow AD-2x=(AB-x\sec\theta)\cos \theta$$ $$\Rightarrow AD=AB\cos \theta+x$$
Next drop $BH \perp AD$ with $H$ on $AD$. In right $ABH$, $AH=AB \cos \theta$. Then $AD=AH+DH$ implies $DH=x$!
Hence in right triangle $BHD$, $BD=2x$, $HD=x=BD/2$. It turns out $\triangle BHD$ is $30^{\circ}-60^{\circ}-90^{\circ}$ and we conclude $$\alpha = 60^{\circ}$$
Here is a purely geometric proof without any trigonometry.
Draw the three lines $a, \, b, \, c$ such that: $$A \, \in \, a \,\,\text{ and } \,\, a \, || \, BC$$ $$B \, \in \, b \,\,\text{ and } \,\, b \, || \, CA$$ $$C \, \in \, c \,\,\text{ and } \,\, c \, || \, AB$$ Furthermore, let $$B^* \, = \, a \, \cap \, c \,\,\, \text{ and } \,\,\, A^* = b \, \cap \, c$$ Then, $ABA^*C$ and $ABCB^*$ are parallelograms and furthermore $$AB = CA^* = CB^*$$ as well as $$AB^* = BC$$ By assumption, $EA = EF$, which means that $\Delta\, EAF$ is isosceles, and consequently $$\angle\, EAF = \angle\, EFA$$ Since $CA^* \,||\, EA$ $$\angle\,CA^*F = \angle\, EAF = \angle\, EFA = \angle\, CFA^*$$ which means that $\Delta\, CA^*F$ is isosceles and $$AB = CF$$ Combining this latter fact with the earlier conclusions, one observes that $$CF = AB = CA^* = CB^*$$ which is possible if and only if triangle $\Delta \, A^*B^*F$ has $90^{\circ}$ angle at vertex $F$, i.e. $\angle \, A^*FB^* = 90^{\circ}$. But that means that: $$\angle \, AFB^* = 180^{\circ} - \angle \, A^*FB^* = 180^{\circ} - 90^{\circ} = 90^{\circ}$$ Recall that by assumption $AF = BD = DC = \frac{1}{2}\,BC$. But by construction, $ABCB^*$ is a parallelogram and $AB^* = BC$ so $$\Delta \, AFB^* \, \text{ is a triangle with the properties that } \, \angle \, AFB^* = 90^{\circ} \,\, \text{ and } \,\, AF = \frac{1}{2}\, AB^* $$
which is possible if and only if $$\angle \, FAB^* = 60^{\circ}$$ However, $AB^*$ is parallel to $BC$ and consequently, by the properties of a pair of parallel lines intersected by a third line, $$\angle \, ADB = \angle \, DAB^* = \angle \, FAB^* = 60^{\circ}$$
