9.4. In triangle $ABC$ with sides $AB=c, BC=a, AC=b$, the median $BM$ is drawn. Incircles are inscribed in triangles $ABM$ and $BCM$. Find the distance between the points of tangency of these incircles with the median $BM$.
Answer: $\quad \frac{|a-c|}{2}$. Solution. Let $P_{1}, Q_{1}, L_{1}$ be the points of tangency of the incircle of triangle $A B M$ with sides $A B, A M$ and $B M$ respectively. Similarly, the points of tangency of the second circle with sides $B C, M C$ and $B M$ are denoted as $P_{2}, Q_{2}, L_{2}$. We need to find $L_{1} L_{2}$. Denote (using the property of the tangent) $A P_{1}=A Q_{1}=x_{1}, \quad M Q_{1}=M L_{1}=y_{1}, B P_{1}=B L_{1}=z_{1}$, $C P_{2}=C Q_{2}=x_{2}, M Q_{2}=M L_{2}=y_{2}, B P_{2}=B L_{2}=z_{2}$. We have two equalities $y_{1}+z_{1}=y_{2}+z_{2}$ (= $B M$ ) and $x_{1}+y_{1}=x_{2}+y_{2}$ (since $M$ is the midpoint of $A C$ ). Adding these equalities, we get $2 y_{1}+\left(x_{1}+z_{1}\right)=2 y_{2}+\left(x_{2}+z_{2}\right)$, i.e., $2 y_{1}+c=2 y_{2}+a$. Therefore, $L_{1} L_{2}=\left|y_{1}-y_{2}\right|=\frac{|a-c|}{2}$.