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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.image

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    It is interesting to note that $BF$ is the bisector of $\angle ABC$ – 2017-02-28
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    @Futurologist question for you! Waiting to see a nice geometric proof! – 2017-02-28
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    *"I already have a proof, but if you can please give another."* You should provide your proof so that potential answerers don't waste their time duplicating your work. (It's really annoying to put work into an answer, only to have the questioner say, "Yeah, I already did it that way.") – 2017-03-03

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$\dfrac{CF}{FA}\cdot\dfrac{EF}{FD} = 2$ 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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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!

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    I upvoted the answer, but have not yet accepted it because I'm still curious about other proofs. Is there a fairly short proof with little to no trigonometry? – 2017-02-27
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    @user_194421 I think, there is this proof. We just need to activate a glande inside the skull. – 2017-02-27
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    I meant a proof without trigonometric functions or with very little of it such as only in the end. The no-trigonometry proof I found was relatively very long. – 2017-02-27
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    @Peter Szilas $\sin\left(90^{\circ}-\frac{\alpha}{2}\right)=\sin\left(90^{\circ}+\frac{\alpha}{2}\right)$ – 2017-03-03