Solutions of triangles $AA_1$, $BB_1$, $CC_1$ are the medians of triangle $ABC$ whose centroid is $G$. If points $A, C_1, G, B_1$ are concylic then prove that $2a^2= b^2 + c^2$.
Thanks 
My try:-
$ar(GBC)=1/3ar(ABC)$
$\frac{1}{2}(GB.GC.\sin(\pi-A))=\frac{1}{3}(\frac{1}{2}bc\sin A)$
$GB.GC=\frac{1}{3}bc$
Now I can't think any further. Here is my diagram:


It is really very simple. Using power of point property
$$BC_{1}.BA=BG.BB_{1}$$
$$\frac{c}{2}.c=\frac{2}{3}m_{b}.m_{b}$$
$$3c^2=4m^2_{b}$$
$$3c^2=2a^2+2c^2-b^2$$
$$2a^2=b^2+c^2$$
