<s>[INST] 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:


*

*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 $$\frac{AB^2}{4}=FA^2=FB^2=FG\cdot FC=\frac{1}{3}FG^2. $$
 [/INST] You were nearly done:
$$
\frac{AB^2}{4}=\frac{1}{3}FC^2 \tag{1}
$$
From Appolonius' Theorem:
$$b^2 + a^2=2 FC^2 + \frac{c^2}{2}$$
$$\implies FC^2=\frac{2a^2+2b^2-c^2}{4}$$
Substituting for $FC^2$ in $Eq(1)$:
$$\frac{c^2}{4}=\frac{1}{3}\frac{2a^2+2b^2-c^2}{4}$$
A little algebra yields the answer.
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