Lemma. $\cos{36} = \frac{1 + \sqrt{5}}{4}$. To prove this we draw a $36$-$72$-$72$ isosceles triangle. Let $A$ be the vertex angle and $B$ and $C$ be the base angles. Draw angle bisector $BD$. Call $BC = a$, $CD = b$, and $AD = c$; then we see that $AB = b + c$. So since $\triangle BDC \sim \triangle ACB$, we have that $\frac{a}{b} = \frac{b + c}{a}$. Also, by the Angle Bisector Theorem, $\frac{a}{b + c} = \frac{b}{c}$, so $\frac{b + c}{a} = \frac{c}{b}$. This means that $\frac{a}{b} = \frac{c}{b}$ if we substitute using our first proportion, so $a = c$. By the Law of Cosines on $\triangle BDC$, we see that $b^2 = 2a^2 - 2a^2\cos{36}$. Also, from the first propotion, $\frac{a}{b} = \frac{a + b}{a}$. So $a^2 = ab + b^2 \Longrightarrow b^2 + ab - a^2 = 0$. Hence, $b = \frac{-a \pm \sqrt{5a^2}}{2} = a \cdot \frac{-1 + \sqrt{5}}{2}$. Now, we plug this into the equation $b^2 = 2a^2 - 2a^2\cos{36}$. $a^2 \cdot \frac{3 - \sqrt{5}}{2} = 2a^2 - 2a^2\cos{36} \Longrightarrow 3 - \sqrt{5} = 4 - 4\cos{36}$. So $\cos{36} = \frac{1 + \sqrt{5}}{4}$. So from here we use Law of Cosines and solve. (I'm figuring out exactly how to proceed from here.)

Let $AD$ intersect $CE$ at $I$, and extend $AD$ to $F$ such that $BACF$ is a rhombus. Then angle chasing, we have $\angle AEI=\angle EAI=\angle IFC=\angle ICF=54$, so $IA=IE$ and $IC=IF\implies CE=AF=2AD=6$.