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Here is the original problem that I was able to solve here:

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: $2BC^{2}=CA^2+AB^2$.

However, this problem has a hidden gem in it. This is, actually, my own conjecture:

If $LM=LN$ then $LM=BC$

I'm calling this "a conjecture" but I guarantee that it's correct. I played with Geogebra a lot and found the statement to be true without doubt (If you see that small dot in the top left corner, that's my dragging point. When moved, it changes the shape of the triangle).

Despite the fact that I was able to solve the original problem, the extended problem refuses to budge. It seems to be much harder than the original one. Or I'm just blind after spending so many hours over the same problem.

enter image description here

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I would say it is obvious.

Just use your graph,denote the mid point of $BC$ as $X$,centroid as $G$.

Fisrt, from $2AX^2+2BX^2=AB^2+AC^2=2BC^2$,and $3GX=AX$, we get $GX ={{\sqrt3}\over 6}BC$.

Second,$LX*XA=BX*XC$,we get $LX={{\sqrt3}\over 6}BC$.

So $CGBL$ is a Parallelogram, so $CL \parallel MB$,and your result follows.

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  • $\begingroup$ You could write some angle chase, I don't see whay the statemnt is true from that paralelogram $\endgroup$ – Aqua Feb 19 at 19:41
  • $\begingroup$ Ok, I see..................... $\endgroup$ – Aqua Feb 19 at 19:45
  • $\begingroup$ @greedoid if parallel, it is either a isosceles trapezium or a rectangle. $\endgroup$ – StAKmod Feb 19 at 19:45
  • $\begingroup$ @greedoid The solution is correct. It follows that $BMCL$ is a trapezoid inscribed in a circle and therefore $BC=LM$ $\endgroup$ – Oldboy Feb 19 at 19:46
  • $\begingroup$ Yes, I already gave him +1 $\endgroup$ – Aqua Feb 19 at 19:46

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