If $|MA|^2=|MB|^2+|MC|^2$. Find the $\angle BMC=?$ My problem is:

The point $M$ is the internal point of the $ABC$ equilateral triangle. Find the $\angle BMC$ if $$|MA|^2=|MB|^2+|MC|^2$$

Is this question related to similar triangles? Or which theorem should I apply?
 A: Hint: 
Rotate $C$ around $B$ for $60^{\circ}$ and let $N$ be a picture of $M$ under this rotation. What can you say for triangles $BMN$ and $AMN$?

Edit: Triangle $BMC$ must be equilateral and since $$ AN^2+MN^2 = CM^2+BM^2 = AM^2$$ we see that triangle $ANM$ is rectangular with 90 at $N$...
A: That equation is essentially the Pythagorean theorem applied to lines MA, MB, MC. Even though they do not form a triangle in this scenario, the equation shows that they can. Line MA is the hypotenuse in this case, which makes angle BMC 90 degrees.
A: Use coordinate geometry. Note that $|MA|^2=|MB|^2+|MC|^2$ is symmetrical in $B\ \&\ C$. This means $M$ is located symmetrically with respect to $B\ \&\ C$. Let the coordinates of $A$ be $(0,\sqrt{3}a)$, $B$ be $(-a,0)$ and $C$ be $(a,0)$. Since $M$ has to be symmetrical, let the coordinates for $M$ be $(0,y)$. Now,
$$|MA|^2=|MB|^2+|MC|^2$$
$$\implies y^2+2\sqrt3ay-a^2=0$$
$$\implies y=(2-\sqrt3)a$$Note that you'll get two values of y but it can't be negative
Now since you've $M$, you can get the angle $BMC$ to be equal to $150$ degrees. I hope you can get there :) 
