[ Intersecting Circles ] [ Auxiliary Similar Triangles ]

A line is drawn through the intersection point of two circles, intersecting the circles again at two points $A$ and $B$.

Find the geometric locus of the midpoints of segments $AB$.
Let $C$ be a common point of two given circles through which the line $A B$ passes, and $D$ be their second common point. Then all triangles $A D B$ are similar to each other.

## Solution

Let $C$ and $D$ be the points of intersection of the given circles, and $M$ be the midpoint of the segment $A B$ passing through point $C$. All triangles $A D B$ are similar to each other by two angles (for example, all angles with vertex $A$ are equal to each other, as they subtend the same arc). Therefore, the angle $C M D$ is equal to one of the angles between $D M$ and $A B$ (depending on which side of $C D$ the point $M$ is located). Consequently, the geometric locus of points is a circle passing through points $C$ and $D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-10.jpg?height=431&width=669&top_left_y=868&top_left_x=703)

Answer

Circle.

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