Proof that $CE=2AD$ I already have a proof, but if you can please give another:

Let $ABC$ be an isosceles triangle with $AB=AC$ and point $D$ be on segment $AB$. The line perpendicular to $AB$ which passes $D$ intersects $BC$ (extended) and $AC$ at $E$ and $F$ respectively. $C$ is on segment $BE$, between $B$ and $E$. If the area of $CEF$ is twice that of $ADF$, prove that $CE=2AD$.

My proof involves the Pythagorean theorem and the similar triangles property.
 A: $\dfrac{CF}{FA}\cdot\dfrac{EF}{FD} = 2$ and $\dfrac{DF}{FE}\cdot \dfrac{EC}{CB}\cdot \dfrac{BA}{AD} = 1.$ This implies that $\dfrac{CE}{AD} = 2\dfrac{BC}{BA}\cdot\dfrac{AF}{FC}$. But $\dfrac{AF}{FC} = \dfrac{AD}{DB}\cdot \dfrac{BE}{CE} = \dfrac{BE}{BD}\cdot\dfrac{AD}{CE} = \dfrac{2AB}{BC}\cdot \dfrac{AD}{CE}$. Multiply these two to get $\dfrac{CE}{AD} = 2\cdot 2\dfrac{AD}{CE}$ or $\dfrac{CE}{AD} = 2.$
Here, the only tricky part was $\dfrac{BE}{BD} = 2\dfrac{AB}{BC}$, which follows from the fact that $\angle BAC = 2\angle BED.$
A: Let $\measuredangle A=\alpha$, $AB=a$ and $AD=xa$.
Hence, $$FC=a-\frac{ax}{\cos\alpha}=\frac{a(\cos\alpha-x)}{\cos\alpha}$$ and
$$CE=BE-BC=\frac{a(1-x)}{\sin\frac{\alpha}{2}}-2a\sin\frac{\alpha}{2}=\frac{a(\cos\alpha-x)}{\sin\frac{\alpha}{2}}.$$
Thus, $$S_{\Delta{FCE}}=\frac{FC\cdot CE\sin\left(90^{\circ}+\frac{\alpha}{2}\right)}{2}=\frac{a^2(\cos\alpha-x)^2\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}\cos\alpha}$$ and
$$S_{\Delta ADF}=\frac{ax\cdot\frac{ax}{\cos\alpha}\sin\alpha}{2}=\frac{a^2x^2\sin\alpha}{2\cos\alpha}.$$
Hence, 
$$\frac{a^2(\cos\alpha-x)^2\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}\cos\alpha}=\frac{a^2x^2\sin\alpha}{\cos\alpha}$$ or
$$(\cos\alpha-x)^2=4x^2\sin^2\frac{\alpha}{2},$$
which says that $$x=\frac{\cos\alpha}{1+2\sin\frac{\alpha}{2}}.$$
Thus, $AD=\frac{a\cos\alpha}{1+2\sin\frac{\alpha}{2}}$,
$$CE=\frac{a\left(\cos\alpha-\frac{\cos\alpha}{1+2\sin\frac{\alpha}{2}}\right)}{\sin\frac{\alpha}{2}}=\frac{2a\cos\alpha}{1+2\sin\frac{\alpha}{2}}$$
and we are done!
