Weird geometry question ($AG^3=CE^2\times AB$) $ABC$ is a triangle with $AB = 2AC$ and $E$ is the midpoint of $AB$. The point $F$ lies on the line $EC$ and the point $G$ lies on the line $BC$ such that $A, F, G$ are collinear and $FG = AC$. Show that $AG^3=CE^2\times AB$

$AC=\frac{1}{2}BC=FG$ therefore $G$ coincides with $B$ and $F$ with $E$ (see the picture below).
From cosine law in triangle $AEC$ we have $CE^2=x^2+x^2-2x^2\cos 2t$
Applying the given relationship $AG^3=CE^2\cdot AB$ we get
$$(2 x)^3= \left(-2 x^2 \cos (2 t)+x^2+x^2\right)(2 x)$$
$$8x^3=8 x^3 \sin ^2 t\to \sin t=1\to t=90°\to 2t=180°$$
which is clearly impossible.
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