<s>[INST] 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.
 [/INST] $\dfrac{CF}{FA}\cdot\dfrac{EF}{FD} = 2$ 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 $\angle BAC = 2\angle BED.$
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