$\underline{\text { Folklore }}$

Consider all triangles $ABC$ where the positions of vertices $B$ and $C$ are fixed, and vertex $A$ moves in the plane of the triangle such that the median $CM$ has the same length. What is the trajectory of point $A$?
In one of the considered triangles, draw a line through vertex $A$ parallel to $C M$, which intersects line $B C$ at point $K$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-35.jpg?height=437&width=797&top_left_y=936&top_left_x=639)

By Thales' theorem, $K C = B C$, so $C M$ is the midline of triangle $A K B$. Therefore, $A K = 2 C M$. Since the positions of vertices $B$ and $C$ are fixed, the position of point $K$ is also fixed. As point $A$ moves, the length of segment $A K$ does not change, so point $A$ moves along a circle with center $K$ and radius $2CM$. (From this circle, two points lying on line $B C$ should be excluded.)

## Answer

Along a circle.

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