10.200. A rhombus with side $m$ is inscribed in a triangle so that one angle is common to both, and the opposite vertex of the rhombus lies on the side of the triangle, dividing this side into segments of length $p$ and $q$. Find the sides of the triangle.
Solution.

In $\triangle A B C$ (Fig. 10.10)

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0716.jpg?height=454&width=687&top_left_y=865&top_left_x=573)

Fig. 10.10, an inscribed rhombus $A D E F$ is given, where $B E=p, E C=q, D E=F E=m$, then $B C=p+q . \triangle D B E \sim \triangle A B C$ (by two angles). Therefore, $\frac{D E}{A C}=\frac{B E}{B C} ; A C=\frac{D E \cdot B C}{B E}=\frac{m(p+q)}{p} . \triangle F E C \sim \triangle A B C$. Therefore, $\frac{F E}{A B}=\frac{E C}{B C} ; A B=\frac{F E \cdot B C}{E C}=\frac{m(p+q)}{q}$. Answer: $p+q ; \frac{m(p+q)}{p} ; \frac{m(p+q)}{q}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0717.jpg?height=467&width=666&top_left_y=156&top_left_x=111)

Fig. 10.11

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0717.jpg?height=597&width=487&top_left_y=61&top_left_x=774)

Fig. 10.12