Regular pentagon and the 42° angle Let ABCDE be a regular pentagon. If $\overline{BF} = \overline{BC}$, calculate $\alpha$.

Using some trigonometry, it's a pretty simple exercise as you can reduce your problem to: $\dfrac{\sin(66°)}{\sin(42°+\frac{\alpha}{2})} = \dfrac{\sin(108°-\alpha)}{\sin(36°+\frac{\alpha}{2})}$.
Sadly, I've been having some problem to prove it geometrically. Any ideas? Thanks in advance.
 A: Let's construct equilateral triangle $ABF$. We have that $\angle{EAF}=48^\circ$, $\angle{AEF}=\frac{1}{2}(180^\circ-48^\circ)=66^\circ$. Thus, $\angle{DEF}=108^\circ-66^\circ=42^\circ$ and $\alpha = \angle{EAF} = \angle{FBC}=48^\circ$

A: 
Arguing in reverse:
Given the regular pentagon $ABCDE$, draw circles with radii $BC=AE=AB$ about centers $B$ and $A$, intersecting at $F$, and
join $EF$, $FA$, and $FB$.
Since $\triangle ABF$ is equilateral, then$$\angle CBF=(108^o-60^o=48^o$$And since $\triangle EFA$ is isosceles, and $\angle FAE=\angle CBF=48^o$, then$$\angle AEF=\angle EFA=66^o$$making$$\angle DEF=(108^o-66^o=42^o$$Conversely then, given $\angle DEF=42^o$ and $BF=BC$, it follows that $\triangle ABF$ is equilateral and $\angle CBF=48^o$.
A: The other answers refer to an equilateral triangle being present in thos construction. Here that advance knowledge is not required.
Draw diagonal $\overline{BE}$, thus completing $\triangle BEF$. From the regularity of the pentagon we know that $\angle BEA$ measures $72°$ and $\angle FEA$ is given to measure $42°$, so the difference $\angle BEF$ measures $30°$. Also the diagonL/side ratio of the pentagon is given by $2\sin54°$, where $\overline{BE}$ is a diagonal and $\overline{FB$ is given to be congruent to the sides.
So the Law of Sines applied to $\triangle BEF$ gives
$\dfrac{\sin\angle EFB}{\sin\angle BEF}=\dfrac{BE}{FB}$
$\dfrac{\sin\angle EFB}{\sin30°}=2\sin54°.$
Plugging $\sin30°=1/2$ then gives
$\sin\angle EFB=\sin 54°,$
but observe that this means $\angle EFB$ could measure either $54°$ or its supplement $126°$. With $F$ lying inside the pentagon and $FB$ congruent to the sides of this pentagon, we must have $|\angle EFB|>72°$ so we select
$\angle EFB|=(180-54)°=126°.$
That leaves $|\angle FBE|=(180-30-126)°=24°$ from which
$|\angle FBC|=(72-24)°=48°.$
We also have
$|\angle ABF|=(108-48)°=60°,$
proving that $\triangle ABF$ is equilateral indeed.
