Belarus — Selection and Training Session

Let $\angle BAC = \alpha$, $\angle CBA = \beta$, $\angle ACB = \gamma$. Let $AH$ be the altitude of the triangle $ABC$ and $X'$ be the point of intersection of $AH$ and the line $MB_1$. Then $\angle X'AB_1 = 90^\circ - \gamma$. Since $\gamma = \angle ACB = \angle CB_1M + \angle B_1MC$ and $CB_1 = CM$ we have $\angle CB_1M = \angle B_1MC = \gamma/2$.

Calculate the value of the angle $\angle B_1X'A$:
$$\begin{align*} \angle B_1X'A &= 180^\circ - \angle X'AB_1 - \angle X'B_1A = [\angle X'B_1A = \angle CB_1M] = \\ &= 180^\circ - (90^\circ - \gamma) - \frac{1}{2}\gamma = 90^\circ + \frac{1}{2}\gamma. \end{align*}$$
By the law of sines for $\triangle B_1X'A$ we obtain
$$\frac{AX'}{\sin \frac{\gamma}{2}} = \frac{AB_1}{\sin (90^\circ + \gamma/2)} = \frac{AB_1}{\cos \frac{\gamma}{2}},$$
i.e. $AX' = AB_1 \tan \frac{\gamma}{2}$. Let $I$ be the center of the incircle $\Gamma$ of the triangle $ABC$ and $\Gamma$ touch the side $BC$ at $A_1$. Let $IA_1 = r$. Since $A_1IC$ is a right-angled triangle we have $IA_1 = CA_1 \tan \frac{\gamma}{2}$. It is well-known fact that $CA_1 = (AC + CB - AB)/2 = AB_1$. Therefore, $AX' = IA_1 = r$.
Similarly, if $X''$ is the point of intersection of $AH$ and the line $LC_1$, we show that $AX'' = r$ which gives the required statement.