ABCDE is a regular pentagon. F is an exterior point (outside the pentagon) such that BFC, BFD, DFA and AFE angles are equal. What is the value of BFC angle?
At this moment I have found that:
- CF=EF (it is possible to draw a circle around a regular pentagon - the tangent segments drawn from one point are equal)
- BF=AF (from BFA triangle which is isosceles)
Also I have some combinations of values for different angles, but nothing that is leading to an answer.
Draw from $C$ the perpendicular to $BF$, to meet line $FD$ at $H$. Line $BF$ is then the perpendicular bisector of $CH$, hence $BH=BC=AB$. It follows that $ABH$ is an equilateral triangle, so that $\angle CBH=108°-60°=48°$. We have then $\angle HBF=24°$ and $\angle FHB=150°$, finally giving the desired result: $\angle HFB=180°-24°-150°=6°$.