Belorusija 2012

We first show that the quadrilateral $A_{1}IM{A}_2$ is cyclic. Indeed, $C_1{B}_1\perp AI$ and $I{C}_1\perp AB$. So, $C_{1}M$ is an altitude in the right-angled triangle $AI{C}_1$, whence $AM\cdot AI=A{C}_1^2$. Further, by the power of a point theorem, $A{C}_1^2=A_{2}A\cdot A{A}_1$. So, $$AM\cdot AI=A{C}_1^2=A_{2}A\cdot A{A}_1,$$ hence $A_{1}IM{A}_2$ is cyclic. Therefore, $\angle IM{A}_1=\angle I{A}_2{A}_1$, $\angle A_{2}MI+\angle I{A}_1{A}_2=180^{\circ }$. Since $I{A}_2=I{A}_1$ we have $\angle I{A}_2{A}_1=\angle IM{A}_1$. Hence, $$\angle QMI=180^{\circ }-\angle A_{2}MI=\angle I{A}_1{A}_2=\angle I{A}_2{A}_1=\angle IM{A}_1.(1)$$ From (1) it follows that the symmetrical image of the line $A_{2}Q$ with respect to the line $AI$ is $P{A}_1$. Hence $P$ is the image of $A_2$ and $Q$ is the image of $A_1$. Therefore the line $PQ$ is the image of $A_2{A}_1$ with respect to the line $AI$, thus the statement follows.