10.178. Find the lengths of the sides of isosceles triangle $ABC$ with base $AC$, if it is known that the lengths of its heights $AN$ and $BM$ are equal to $\boldsymbol{n}$ and $m$ respectively.
Solution.

$\triangle C M B \sim \triangle C N A$ (Fig. 10.147), from which $\frac{B M}{A N}=\frac{B C}{A C}$. Therefore, $A C=\frac{n}{m} B C$, $M C=\frac{1}{2} A C=\frac{n}{2 m} B C$. From $\triangle B M C$, $B C^{2}=M B^{2}+M C^{2}$, $B C^{2}=m^{2}+B C^{2}\left(\frac{n}{2 m}\right)^{2}$, from which $B C=\frac{2 m^{2}}{\sqrt{4 m^{2}-n^{2}}}$, $A C=\frac{2 m n}{\sqrt{4 m^{2}-n^{2}}}$.

Answer: $\frac{2 m^{2}}{\sqrt{4 m^{2}-n^{2}}}, \frac{2 m n}{\sqrt{4 m^{2}-n^{2}}}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-639.jpg?height=370&width=550&top_left_y=91&top_left_x=107)

Fig. 10.148

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-639.jpg?height=386&width=536&top_left_y=91&top_left_x=700)

Fig. 10.149