Let $ABCDE$ be a regular pentagon, and $F$ some point on the arc $AB$ of the circumcircle of $ABCDE$. Show that \[ \frac{FD}{FE + FC} = \frac{FB + FA}{FD} = \frac{-1 + \sqrt{5}}{2}, \]and that $FD + FB + FA = FE + FC$.
Source: 2017 Pan-African Shortlist - G3
Tags: geometry, pentagon, circumcircle
Let $ABCDE$ be a regular pentagon, and $F$ some point on the arc $AB$ of the circumcircle of $ABCDE$. Show that \[ \frac{FD}{FE + FC} = \frac{FB + FA}{FD} = \frac{-1 + \sqrt{5}}{2}, \]and that $FD + FB + FA = FE + FC$.