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.
Best Answer
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 $$\alpha = 60^{\circ}$$