Find the geometric locus of points $M$, lying inside the equilateral triangle $A B C$, for which $M A^{2}=$ $M B^{2}+M C^{2}$

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When rotating by $60^{\circ}$ with center $A$, which maps $B$ to $C$, point $M$ transitions to some point $M^{\prime}$, and point $C$ to point $D$. The equality $M A^{2}=M B^{2}+M C^{2}$ is equivalent to the equality $M^{\prime} M^{2}=M^{\prime} C^{2}+M C^{2}$, i.e., that $\angle M C M^{\prime}=$ $90^{\circ}$, which means $\angle M C B+\angle M B C=\angle M C B+\angle M^{\prime} C D=120^{\circ}-90^{\circ}=30^{\circ}$, i.e., $\angle B M C=150^{\circ}$. The desired $Г М Т$ - an arc of a circle lying inside the triangle, from which the segment $B C$ is seen at an angle of $150^{\circ}$.