Geometry - prove, that the center of circumscribed circle of a triangle lays on line.

Question

Inside the angle, which vertice is the point $M$, the randomly selected point $A$ is drawn. From this point the ball is released, which at first reflected from one side of the angle at point $B$, after that it reflected from another angle's side at point $C$ and came back to point $A$ ("the angle of incidence equals the angle of reflection"). Prove, that the center of circumscribed circle of a triangle $BCM$ lays on line $AM$.

Answer 1

By the reflection assumption, $BM$ is the external angle bisector of $\angle ABC$, and $CM$ is the external bisector of $\angle ACB$. Therefore $M$ is an excentre of $\triangle ABC$, and $AM$ is the bisector of $\angle BAC$. Let $P$ be the incentre of $\triangle ABC$. Then the internal bisectors $BP$ and $CP$ are respectively perpendicular to the external bisectors $BM$ and $CM$, so $\triangle PBM$ and $\triangle PCM$ are right triangles. Therefore $BPCM$ are concyclic and $MP$ is a diameter of the circle. In particular, $AM$ passes through the centre of the circle.