2016 Oral Moscow Geometry Olympiad P1 wrote: The line passing through the center $I$ of the inscribed circle of a triangle $ABC$, perpendicular to $AI$ and intersects sides $AB$ and $AC$ at points $C'$ and $B'$, respectively. In the triangles $BC'I$ and $CB'I$, the heights $C'C_1$ and $B'B_1$ were drawn, respectively. Prove that the middle of the segment $B_1C_1$ lies on a straight line passing through point $I$ and perpendicular to $BC$. Let $ID\cap B_1C_1=T$. First of all notice that by congruency we get that $IC'=IB'$ and $\angle C'IB=\frac{\angle C}{2}$ and $\angle B_1IB'=\frac{\angle B}{2}$. So, $\triangle C'C_1I\sim\triangle IDC$ and $\triangle B'B_1I\sim\triangle IDB$. Now by Ratio Lemma we get that $$\frac{C_1T}{B_1T}=\frac{C_1I\cdot \sin \angle BID}{B_1I\cdot \sin \angle CID}=\frac{C_1I\cdot\sin\angle IB'B_1}{B_1I\cdot\sin\angle IC'C_1}=\frac{C_1I\cdot\frac{IB_1}{B'I}}{B_1I\cdot\frac{IC_1}{IC'}}=1\implies B_1T=C_1T.\qquad \blacksquare $$