Iranian Mathematical Olympiad

First, note that the circumcircles of triangles $AI{I}_1$ and $AI{I}_2$ are perpendicular, because $$\angle A{I}_{1}I+\angle A{I}_{2}I=\frac{\angle B+\angle BAD}{2}+\frac{\angle C+\angle CAD}{2}=90^{\circ }$$ Under an inversion with center $A$ and power $AB\times AC$ and then a reflection with respect to the bisector of $\angle BAC$, pairs $(B,C)$ and $(I,I_a)$ are mapped to each other. Let $M^{\prime }$ and $N^{\prime }$ be the image of $M$ and $N$ under this transformation, respectively. Circumcircles of $A{I}_{1}I$ and $A{I}_{2}I$ (which pass through the center of inversion) are mapped to lines $I_a{M}^{\prime }$ and $I_a{N}^{\prime }$. It is obvious that in this transformation, the circumcircle of triangle $ABC$ and the line $BC$ are mapped to each other, so $M^{\prime }$ and $N^{\prime }$ are on the line $BC$. Now, the assertion becomes equivalent to proving that the circumcircle of $A{M}^{\prime }{N}^{\prime }$ passes through some constant point other than $A$ as $D$ varies.

Since inversion and reflection both preserve the angles, we have $I_a{M}^{\prime }\perp I_a{N}^{\prime }$. If $X$ is the foot of the perpendicular line from $I_a$ to $BC$, then the power of $X$ with respect to this circle is $M^{\prime }X\cdot N^{\prime }X=I_a{X}^2$ which is constant as $D$ varies. Therefore, the power of $X$ with respect to the circumcircle of $A{M}^{\prime }{N}^{\prime }$ is constant. Hence $X$ lies on the radical axis of these circles and another point on the line $AX$ is the common point of all these circles.