Find angle $BFE$ in the figure

Question

In the figure, $ABCDE$ is a regular pentagon and $\angle FBC = 90^\circ$ and $DC=FE$, find $\angle BFE$.

Answer 1

Let $AB=1$.

Thus, by law if sines for $\Delta FEB$ we obtain: $$\frac{\sin\measuredangle BFE}{BE}=\frac{\sin\measuredangle FBE}{FE}$$ or $$\frac{\sin\measuredangle BFE}{2\sin54^{\circ}}=\frac{\sin18^{\circ}}{1}.$$ Id est, $$\sin\measuredangle BFE=2\sin54^{\circ}\sin18^{\circ}=2\cos36^{\circ}\cos72^{\circ}=$$ $$=\frac{4\sin36^{\circ}\cos36^{\circ}\cos72^{\circ}}{2\sin36^{\circ}}=\frac{\sin144^{\circ}}{2\sin36^{\circ}}=\frac{1}{2}$$ and since $\angle BFE$ is an acute angle, we obtain $$\measuredangle BFE=30^{\circ}.$$ Done!

Answer 2

Using only elementary geometry, in the given figure, join $EB$. From the nature of the regular pentagon, $\angle ABE=36^o$. Through $E$ draw $EH$ perpendicular to $FB$, meeting $BA$ extended at $K$. Since $EK$ is parallel to $CB$, then $\angle EKA$, supplementary to $\angle ABC,=72^o$. But also, $\angle KAE$, supplementary to $\angle EAB$,$=72^o$. Therefore $\triangle EKA$ is isosceles, with $EK=EA$. Further, since triangles $EKA$ and $EKB$ are similar, then $BE=BK$. And since $BH$ is perpendicular to $EK$, then $EH=HK$, making $$EH=\frac12EK$$But $EK=EA=EF$. Therefore $$EH=\frac12EF$$And since $EF$ is the hypotenuse of right triangle $EHF$, then$$\angle EFH=30^o$$