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 $AG^3=CE^2\times AB$
 A: $AC=\frac{1}{2}BC=FG$ 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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A: The claim is true. However, in order to have a $F \ne E$ so that $FG = 1$, one need to look for a $F$ outside the line segment $CE$ (more precisely, on the ray start at $C$ pointing away from $E$).

For simplicity, we will only consider the case $AC = AE = EB = FG = 1$.
For any point $P$ in the plane, let $z_P = (u_P,v_P,w_P)$ be its barycentric coordinates with respect to $\triangle ABC$. i.e $u_P, v_P, w_P$ are the three real numbers such that:
$$\vec{P} = u_P \vec{A} + v_P \vec{B} + w_P\vec{C}\quad\text{ and } \quad u_P + v_P + w_P = 1$$
Since $E$ is mid-point of $AB$, we have $z_E = (\frac12,\frac12,0)$.
Since $F$ lies on $CE$, $$z_F = (-s,-s,1+2s)$$ for some number $s$. In order to met the condition $FG = 1$ for a $F \ne E$, we need $u_F$ to be negative which is equivalent to $s$ to be positive!
Since $G$ lies on $BC$, $u_G = 0$. Together with the fact $G$ lies on $AF$,
we find $$z_G = \left(0,-\frac{s}{1+s},\frac{1+2s}{1+s}\right)$$
Notice $\displaystyle\;\frac{u_F - u_A}{u_G - u_A} = \frac{-s-1}{-1} = s+1$. We have
$$AF = (1+s)AG \implies FG = AF-AG = sAG$$
If $FG = 1$, we will have $$AG = \frac1s\quad\text{ and }\quad AF = 1 + \frac1s$$
To proceed, we need another formula to compute $AF$.
Let $\alpha = \frac12\angle BAC$, $\beta = CE = 2\sin\alpha$ and $\delta = AG = \frac1s$.
Since $AC = AE$, if we let $M$ be the mid-point of $EC$, $\triangle AMF$ is a right angled triangle. It is easy to see
$FM : CM = s + \frac14 : \frac14$, this implies
$$AF^2 = AM^2 + MF^2 = \cos^2\alpha + \sin^2\alpha(4s+1)^2
= 1 + 2\beta^2 s(2s+1)
$$
Combine with $AF = FG + AG = 1 + \delta$ and $s = \frac1{\delta}$, we obtain
$$(1+\delta)^2 = AF^2 = 1 + 2\beta^2 \frac{\delta+2}{\delta^2}
\quad\implies\quad \delta^3 = 2\beta^2$$
The last equality is precisely what we want to show:
$$AG^3 = \delta^3 = 2\beta^2 = AB(CE)^2$$
