18.15. Find the geometric locus of points $M$, lying inside the equilateral triangle $ABC$, for which $M A^{2}=M B^{2}+M C^{2}$.
18.15. 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 locus of points is the arc of a circle lying inside the triangle from which the segment $B C$ is seen at an angle of $150^{\circ}$.