Let $G$ -intersection $CF$ and $EA$. Then $\angle GAF=72,\angle FCB=18, \angle CFB=54 \to AG=AF$. $EC \parallel AF$ so $EG=AG+AE=AE+AF=EC=EB$

Slightly different... Draw a point $F'$ on ray $EA$ such that $AF'=AF$. Some easy angle chasing shows that $\angle AFF'+\angle CFA=54+126=180$. Hence, $F,F'$ and $C$ are collinear. Angle chasing shows $\angle ECF=54=\angle EFC$. So, $EA+AF=AE+AF'=EF'=EC=EB$ as needed.

Let $x$ the measure of the side of $ABCDE$, consider $G$ the intersection of the parallel $DC$ passing by $F$ with $AE$, let $AF=y$, we are going to calculate $BE$ and $y$ in terms of $x$ Note $AF=AG=y$ and $GF=x$, then is easily to check $x=GF=2\cdot \sin54\cdot y \Leftrightarrow AF=y=\frac{x}{2\sin54}$, and $BE=2\cdot \sin54 \cdot x$ Then $AE+AF=BE \Leftrightarrow$ $x+y=2\cdot \sin54 \cdot x \Leftrightarrow$ $x+\frac{x}{2\sin54}=2\cdot \sin54 \cdot x \Leftrightarrow$ $1+\frac{1}{2\sin54}=2\sin 54 \Leftrightarrow$ $2\sin54=\frac{1+\sqrt{5}}{2}$ Which is well-know. $\blacksquare$