10.004. The height of the rhombus, drawn from the vertex of the obtuse angle, divides its side into segments of length $m$ and $n$. Determine the diagonals of the rhombus.
Solution.

The length of the side of the rhombus is $m+n$. From $\triangle A B K$ (Fig. 10.4) we find $B K^{2}=(m+n)^{2}-m^{2}$. In $\triangle B K D$ we have

$$
B D^{2}=B K^{2}+n^{2}=(m+n)^{2}-m^{2}+n^{2}=2 n(m+n)
$$

i.e., $B D=\sqrt{2 n(m+n)}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-558.jpg?height=400&width=384&top_left_y=92&top_left_x=160)

Fig. 10.5

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-558.jpg?height=418&width=534&top_left_y=91&top_left_x=701)

Fig. 10.6

Since $A C^{2}+B D^{2}=4 A D^{2}=4(m+n)^{2}$, then

$$
A C^{2}=4(m+n)^{2}-2 m n-2 n^{2}=4 m^{2}+6 m n+2 n^{2},
$$

i.e., $A C=\sqrt{4 m^{2}+6 m n+2 n^{2}}$.

Answer: $B D=\sqrt{2 n(m+n)}, A C=\sqrt{4 m^{2}+6 m n+2 n^{2}}$.