Let $ABC$ be a triangle with unequal sides. The medians of $ABC$,
when extended, intersect its circumcircle in points $L, M, N$. If $L$ lies
on the median through $A$ and $LM = LN$, prove that: $2BC^{2}=CA^2+AB^2$.
So I first wrote midpoint theorem $AB^2+AC^2=2AX^2 +2XB^2$ (refer to the diagram) and so I tried to put $AX$ in terms of $BX$ or $BC$ but I can't find a way. I'm also not sure where $LM$ and $LN$ come in, so it would be helpful if someone could tell me the line of thought. Thanks!
On
Below is a brute force solution. Though there should be more elegant solutions, I have decided to present this one due to the beauty of some resulting algebraic expressions.
First, it should be clarified that (as will be seen) the real meaning of the condition "a triangle with unequal sides" is "a triangle with $AC\ne AB$".
Everywhere below we use notation: $$ BC=a,\quad CA=b,\quad AB=c,\quad AX=m_a,\quad BY=m_b,\quad CZ=m_c. $$
First we compute the length of the segment $LM$. For this we need the lengths $OL$, $OM$ and cosine of the angle $\widehat{LOM}$: $$ OL=OX+XL=\frac{m_a}3+\frac{a^2}{4m_a}=\frac{4m_a^2+3a^2}{12m_a}\\ =\frac{(2b^2+2c^2-a^2)+3a^2}{12m_a}=\frac{a^2+b^2+c^2}{6m_a}, $$ and similarly $$ OM=\frac{a^2+b^2+c^2}{6m_b}. $$ Further: $$ \cos \widehat{LOM}=\frac{OA^2+OB^2-BC^2}{2OA\cdot OB}=\frac{m_a^2+m_b^2-\left(\frac32c\right)^2}{2m_am_b} $$ Combining together one obtains: $$ LM=\frac{a^2+b^2+c^2}{6}\left[\frac1{m_a^2}+\frac1{m_b^2} -\frac2{m_am_b}\frac{m_a^2+m_b^2-\left(\frac32c\right)^2}{2m_am_b}\right]^{1/2}\\ =\frac{(a^2+b^2+c^2)c}{4m_am_b}. $$
Similarly: $$ LN=\frac{(a^2+b^2+c^2)b}{4m_am_c}. $$
Combining two last equalities one obtains: $$ LM=LN\implies m_bb=m_cc\\ \implies (2a^2+2c^2-b^2)b^2=(2a^2+2b^2-c^2)c^2\implies (2a^2-b^2-c^2)(b^2-c^2)=0. $$ As $b\ne c$ one is left with $$ b^2+c^2=2a^2. $$
The equality $ m_bb=m_cc$ has a simple geometric interpretation. It means that the medians $BY$ and $CZ$ intersect the respective sides $AC$ and $AB$ at equal angles. If $\angle BYC=\angle CZB$ then $b=c$. If $\angle BYC=\angle CZA$ then $b^2+c^2=2a^2$.
The required triangle $ABC$ can be constructed as follows:
Draw a circle $\cal C$ with radius $\frac{\sqrt3}2 BC$ centered at the midpoint $X$ of the segment $BC$. Then arbitrary point of the circle (except for the intersection points with $BC$) can be taken as the third vertex $A$ of the triangle.
Notice that triangles $LMG\sim BAG$ and $LNG\sim CAG$ are similar.
From the first pair of similar triangles we have:
$$\frac{BA}{BG}=\frac{LM}{LG}=\frac{LN}{LG}\tag{1}$$
From the second pair of similar triangles we have:
$$\frac{LN}{LG}=\frac{CA}{CG}\tag{2}$$
From (1) and (2):
$$\frac{BA}{BG}=\frac{CA}{CG}$$
$${BA}^2\cdot{CG}^2={CA}^2\cdot{BG}^2\tag{3}$$
Use the fact that:
$$CG^2=\frac19(2a^2+2b^2-c^2)\tag{4}$$
$$BG^2=\frac19(2a^2+2c^2-b^2)\tag{5}$$
$$BA=c,\ \ CA=b\tag{6}$$
Replace (4), (5), (6) into (3) and you get, after some simplifaction:
$$(c^2-b^2)(c^2+b^2-2a^2)=0$$
Triangle is scalene so $b\ne c$. It follows that:
$$b^2+c^2=2a^2$$