Given a triangle $ABC$. It is known that $AB=4$, $AC=2$, and $BC=3$. The bisector of angle $BAC$ intersects side $BC$ at point $K$. A line passing through point $B$ parallel to $AC$ intersects the extension of the bisector $AK$ at point $M$. Find $KM$.

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By the property of the bisector of a triangle $B K: K C=A B: A C=2: 1$. Therefore, $B K=2, K C=1$. According to problem $\underline{53122} A K^{2}=A B \cdot A C-B K \cdot K C=6$.

Since triangles АСК and МВК are similar by two angles, $K M=2 A K$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-36.jpg?height=292&width=512&top_left_y=1506&top_left_x=773)

## Answer

$2 \sqrt{6}$.