<s>[INST] Can we prove this geometry question without too much calculation? Can we prove the following geometry question without too much calculation?
$ABCD$ is a square, $AC=AE$, $DF=DE$, $AC$ and $BF$ meet at $G$, $BG=CE$. 

Prove: $BG \parallel CE$.


 [/INST] It looks to be related with a dissection of a dodecagon in three squares:

Let we assume that the square has a unit side, $\widehat{CAE}=\theta$ and $H$ is the projection of $D$ on $AE$.
If we prove $\theta=30^\circ$, the claim easily follows. We have $AE=\sqrt{2}$ and
$$ AH = \cos\left(\frac{\pi}{4}-\theta\right) = \frac{\cos\theta+\sin\theta}{\sqrt{2}}, $$
so $AF=\sqrt{2}(\cos\theta+\sin\theta-1)$. By the cosine theorem
$$ BF^2 = 1+AF^2-2 AF\cos\left(\theta+\frac{\pi}{4}\right) $$
and
$$ \frac{BG}{GF}=\frac{[BAG]}{[GAF]}=\frac{\sin\frac{\pi}{4}}{AF\sin\theta}, $$
hence
$$ BG^2 = \left(\frac{\sin\frac{\pi}{4}}{\sin\frac{\pi}{4}+AF\sin\theta}\right)^2 BF^2 = \frac{5-2 \cos\theta-2\cos(2\theta)-6\sin\theta+2\sin(2\theta)}{(2-\cos(2\theta)-2\sin(\theta)+\sin(2\theta))^2}$$
and by equating $BG^2$ and $CE^2=4(1-\cos\theta)$
$$ \sin\theta=\frac{1}{2} $$
follows (not easily - better to have the help of a CAS).
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