10.377 The bisector of a triangle's angle divides the opposite side into segments of length 4 and 2 cm, and the height drawn to the same side is $\sqrt{15}$ cm. What are the lengths of the sides of the triangle, given that they are expressed as integers?
Solution. Let the bisector of angle $A$ divide side $B C$ of the given $\triangle A B C$ into segments 2 and 4. Suppose $A Bh=\sqrt{15}(h$ - the height of $\triangle A B C) \Rightarrow \sqrt{15}<n<6 \Rightarrow 4 \leq n \leq 5$.

If $n=5$, then by Heron's formula $S_{A B C}=\sqrt{\frac{21}{2} \cdot \frac{11}{2} \cdot \frac{9}{2} \cdot \frac{1}{2}}=\frac{3 \sqrt{231}}{4}$.

But $S_{A B C}=B C \cdot \frac{h}{2}=3 \sqrt{15}$, which is impossible. Therefore, $n=4$.

Answer: 4 cm, 8 cm, 6 cm.