Golden ratio in regular pentagon

Question

Let $ABCDE$ be a regular pentagon, F in arc small BA. Show that $\frac{FD}{FE+FC}=\frac{FB+FA}{FD}=0.618033$ the golden ratio; and $FD+FB+FA=FE+FC$

Answer 1

Using Ptolemy theorem for a quadrilateral $CDEF$ we get $$FE \cdot CD + FC \cdot DE = FD \cdot EC,$$ therefore $$\frac{FD}{FE+FC} = \frac{DE}{EC} = \frac{1}{\varphi}.$$

Analogously, Ptolemy theorem for $BDAF$ yields $$FA \cdot BD + FB \cdot AD = FD \cdot AB$$ which implies $$\frac{FB+FA}{FD} = \frac{AB}{DB} = \frac{1}{\varphi}.$$

Using the results above we get $FB+FA=\frac{FD}{\varphi}$ and $FE+FC=\varphi\cdot FD$. Keeping in mind that $\varphi+1=\varphi^2$, we get

$$FD+FB+FA = FD+\frac{FD}{\varphi} = \frac{\varphi+1}{\varphi}\cdot FD = \frac{\varphi^2}{\varphi}\cdot FD = \varphi \cdot FD = FE + FC.$$