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


Let $\measuredangle LAC=\alpha_1$, $\measuredangle LAB=\alpha_2$, $\measuredangle MBA=\beta_1$, $\measuredangle MBC=\beta_2$, $\measuredangle NCB=\gamma_1$ and $\measuredangle NCA=\gamma_2$.
Thus,
$$\gamma_1+\alpha_2=\measuredangle NBM+\measuredangle BML=\measuredangle NML=\measuredangle MNL=\measuredangle MNC+\measuredangle LNC=\beta_2+\alpha_1,$$
which gives
$$\cos(\gamma_1+\alpha_2)=\cos(\beta_2+\alpha_1)$$ or 
$$\cos\gamma_1\cos\alpha_2-\sin\gamma_1\sin\alpha_2=\cos\beta_2\cos\alpha_1-\sin\beta_2\sin\alpha_1$$ or by law of cosines and by law of sines
$$\frac{a^2+m_c^2-\frac{c^2}{4}}{2am_c}\cdot\frac{c^2+m_a^2-\frac{a^2}{4}}{2cm_a}-\frac{\frac{c}{2}\sin\beta}{m_c}\cdot\frac{\frac{a}{2}\sin\beta}{m_a}=$$
$$=\frac{a^2+m_b^2-\frac{b^2}{4}}{2am_b}\cdot\frac{b^2+m_a^2-\frac{a^2}{4}}{2bm_a}-\frac{\frac{b}{2}\sin\gamma}{m_b}\cdot\frac{\frac{a}{2}\sin\gamma}{m_a}$$ or
$$\frac{\left(a^2+\frac{1}{4}(2a^2+2b^2-c^2)-\frac{c^2}{4}\right)\left(c^2+\frac{1}{4}(2b^2+2c^2-a^2)-\frac{a^2}{4}\right)}{acm_c}-\frac{ac\sin^2\beta}{m_c}=$$
$$=\frac{\left(a^2+\frac{1}{4}(2a^2+2c^2-b^2)-\frac{b^2}{4}\right)\left(b^2+\frac{1}{4}(2b^2+2c^2-a^2)-\frac{a^2}{4}\right)}{abm_b}-\frac{ab\sin^2\gamma}{m_b}$$ or
$$\frac{(3a^2+b^2-c^2)(3c^2+b^2-a^2)}{4acm_c}-\frac{ac\left(\frac{2S}{ac}\right)^2}{m_c}=$$
$$=\frac{(3a^2+c^2-b^2)(3b^2+c^2-a^2)}{4abm_b}-\frac{ab\left(\frac{2S}{ab}\right)^2}{m_b}$$ or 
$$\frac{(3a^2+b^2-c^2)(3c^2+b^2-a^2)-\sum\limits_{cyc}(2a^2b^2-a^4)}{cm_c}=$$
$$=\frac{(3a^2+c^2-b^2)(3b^2+c^2-a^2)-\sum\limits_{cyc}(2a^2b^2-a^4)}{bm_b}$$ or
$$\frac{a^4+c^4-b^4-4a^2c^2}{c\sqrt{2a^2+2b^2-c^2}}=\frac{a^4+b^4-c^4-4a^2b^2}{b\sqrt{2a^2+2c^2-b^2}}$$ or
$$(a^2+b^2+c^2)^2(b^2+c^2-2a^2)(b^2-c^2)(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$ and we are done!
