Property of Medians and Cirumcircle

Question

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

Answer 1

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 $$BG.GM=CG.GN$$

$$\Rightarrow \frac{2}{3}m_{b}.GM=\frac{2}{3}m_{c}.NG$$

$$\Rightarrow \frac{2}{3}m_{b}.\frac{b}{c}.NG=\frac{2}{3}m_{c}.NG$$

$$\Rightarrow b.m_{b}=c.m_{c}$$

$$\Rightarrow b\sqrt{2c^2+2a^2-b^2}=c\sqrt{2a^2+2b^2-c^2}$$

$$\Rightarrow 2b^2c^2+2a^2b^2-b^4=2a^2c^2+2b^2c^2-c^4$$

$$\Rightarrow 2a^2(b^2-c^2)=b^4-c^4$$

$$\Rightarrow 2a^2=b^2+c^2$$

Answer 2

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!