<s>[INST] Property of Medians and Cirumcircle Let $ABC$ be a non-isosceles triangle. Medians of $\triangle ABC$ intersect the circumcircle in points $L,M,N$. If $L$ lies on the median of $BC$ and $LM=LN$, then prove that $2a^2=b^2+c^2$.
My Attempt:

Let $G$ be the centroid of $\triangle ABC$ and $D$ be the mid-point of $BC$.
Since $LM=LN$, therefore $LM$ and $LN$ will subtend equal angles on the circumference of circumcircle
$\Rightarrow \angle GBL=\angle LCG$
$GL=GL$
Altitude of $\triangle BGL$ from $B$ to $GL$=Altitude of $\triangle LGC$ from $C$ to $GL$
Area($\triangle BGL$)=Area($\triangle LGC$)
Now, based on this ,can it be said that $\square GBLC$ is a parallelogram. If yes then 
$$DL=GD=\frac{m_{a}}{3}$$
$\Rightarrow AD.DG=BD.DC$
$$m_{a}.\frac{m_{a}}{3}=\frac{a^2}{4}$$
$\Rightarrow 4m^2_{a}=3a^2$
$\Rightarrow 2b^2+2c^2-a^2=3a^2$
$\Rightarrow b^2+c^2=2a^2$
I am not sure whether this justification is sufficient(the one that I have written in bold). What more can be added to seal the issue

 [/INST] 
Consider $\triangle AGB$ and $\triangle MGL$
$$\angle BGA=\angle LGM$$
$$\angle GAB=\angle GML$$
$$\triangle AGB \sim \triangle MGL$$
$$\frac{GB}{GL}=\frac{AG}{GM}=\frac{AB}{LM}$$
$$AG=\frac{c}{LM}GM$$

Similarly $\triangle AGC \sim \triangle NGL$
$$\frac{AG}{GN}=\frac{GC}{LG}=\frac{AC}{LN}$$
$$AG=\frac{b}{LN}GN$$
Thus$$AG=\frac{b}{LN}GN=\frac{c}{LM}GM$$
$$\Rightarrow c.GM=b.NG$$
Also, we have $$BG.GM=CG.GN$$
$$\Rightarrow \frac{2}{3}m_{b}.GM=\frac{2}{3}m_{c}.NG$$
$$\Rightarrow \frac{2}{3}m_{b}.\frac{b}{c}.NG=\frac{2}{3}m_{c}.NG$$
$$\Rightarrow b.m_{b}=c.m_{c}$$
$$\Rightarrow b\sqrt{2c^2+2a^2-b^2}=c\sqrt{2a^2+2b^2-c^2}$$
$$\Rightarrow 2b^2c^2+2a^2b^2-b^4=2a^2c^2+2b^2c^2-c^4$$
$$\Rightarrow 2a^2(b^2-c^2)=b^4-c^4$$
$$\Rightarrow 2a^2=b^2+c^2$$
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