Given a triangle with sides $A B=2, B C=3, A C=4$. A circle is inscribed in it, and the point $M$ where the circle touches side $B C$ is connected to point $A$. Circles are inscribed in triangles $A M B$ and $A M C$. Find the distance between the points where these circles touch the line $A M$.
Since $K$ is the point of tangency of the incircle inscribed in triangle $AMB$ with side $AM$, $AK = (AB + AM - BM) / 2$.

Similarly, expressing $AL$ and subtracting the obtained expressions from each other, we get

$KL = (AB - AC - BM + MC) / 2$.

| 3 Topics: | $\left[\begin{array}{lc}\text { [ Cube } & \text { [ } \\ \text { [ Unfolding helps solve the problem }\end{array}\right]$ | Difficulty: 3+ |
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Substituting the expression $BM = (BC + AB - AC) / 2$ into this equation, we get $KL = 0$.

Answer: 0.