Centroid and midpoints cyclic then prove that $AC^2 + BC^2 = 2AB^2.$ 
Let $ABC$ be a non-equilateral triangle with integer sides. Let $D$ and $E$ be respectively the mid-points of $BC$ and $CA$ ; let $G$ be the centroid of $\Delta{ABC}$. Suppose, $D$, $C$, $E$, $G$ are concyclic.Prove that  $AC^2 + BC^2 = 2AB^2.$

My progress:


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*By angle chase we get, $$\angle GBA=\angle GED=\angle GCD\implies (CGB)~~\text{is tangent to}~~AB.$$

*Similarly, we get $$\angle GAB=\angle GDE=\angle GCE\implies (CGA)~~\text{is tangent to}~~AB.$$

*By power of point, we have $$\frac{AB^2}{4}=FA^2=FB^2=FG\cdot FC=\frac{1}{3}FG^2. $$

From your working, $ \displaystyle \frac{AB^2}{4}=FA^2=FB^2=FG\cdot FC=\frac{1}{3}FC^2$
Now to finish off, you can use the formula of median in terms of side lengths.
$FC^2 = \cfrac{1}{4} (2 AC^2 + 2 BC^2 - AB^2)$
So, $\cfrac{AB^2}{4} = \cfrac{1}{12} (2 AC^2 + 2 BC^2 - AB^2)$
$ \implies 2 AB^2 = AC^2 + BC^2$
