Selection and Training Session

Let $\angle BAC=\alpha$, $\angle CBA=\beta$, $\angle ACB=\gamma$. Let $AH$ be the altitude of the triangle $ABC$ and $X^{\prime }$ be the point of intersection of $AH$ and the line $M{B}_1$. Then $\angle X^{\prime }A{B}_1=90^{\circ }-\gamma$. Since $\gamma =\angle ACB=\angle C{B}_{1}M+\angle B_{1}MC$ and $C{B}_1=CM$ we have $\angle C{B}_{1}M=\angle B_{1}MC=\gamma /2$.



Calculate the value of the angle $\angle B_1{X}^{\prime }A$: $$\begin{array}{rl}\angle B_1{X}^{\prime }A& =180^{\circ }-\angle X^{\prime }A{B}_1-\angle X^{\prime }{B}_{1}A=[\angle X^{\prime }{B}_{1}A=\angle C{B}_{1}M]=\\ & =180^{\circ }-(90^{\circ }-\gamma )-\frac{1}{2}\gamma =90^{\circ }+\frac{1}{2}\gamma .\end{array}$$ By the law of sines for $△B_1{X}^{\prime }A$ we obtain $$\frac{A{X}^{\prime }}{\sin \frac{\gamma }{2}}=\frac{A{B}_1}{\sin (90^{\circ }+\gamma /2)}=\frac{A{B}_1}{\cos \frac{\gamma }{2}},$$ i.e. $A{X}^{\prime }=A{B}_1\tan \frac{\gamma }{2}$. Let $I$ be the center of the incircle $\Gamma$ of the triangle $ABC$ and $\Gamma$ touch the side $BC$ at $A_1$. Let $I{A}_1=r$. Since $A_{1}IC$ is a right-angled triangle we have $I{A}_1=C{A}_1\tan \frac{\gamma }{2}$. It is well-known fact that $C{A}_1=(AC+CB-AB)/2=A{B}_1$. Therefore, $A{X}^{\prime }=I{A}_1=r$. Similarly, if $X^{\prime \prime }$ is the point of intersection of $AH$ and the line $L{C}_1$, we show that $A{X}^{\prime \prime }=r$ which gives the required statement.