Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it

Let $ABC$ be a triangle, and $M$ the midpoint of $[BC]$ . We denote $I_{b}$ and $I_{c}$ as the centers of the incircles of $AMB$ and $AMC$ . Show that the second point of intersection of the circumcircles of triangles $ABI_{b}$ and $ACI_{c}$ lies on the line $(AM)$ .

Solution

Let $T$ be the intersection between the circle with diameter $[B C]$ and the ray $\left[M A\right)$ . It suffices to show that $T$ lies on the circle $A B I_{b}$ (by symmetry of the roles of $B$ and $C$ , this will show that it also lies on the circle $A C I_{c}$ ).
Since $\widehat{B T C}$ is a right angle, we have $\widehat{A T B}=90^{\circ}+\frac{1}{2} \widehat{A M B}=\widehat{A I_{b} B}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.