In a trapezoid, the midline is equal to 7, the height is equal to $\frac{15 \sqrt{3}}{7}$, and the angle between the diagonals opposite the base is $120^{\circ}$.

Find the diagonals of the trapezoid.
Through the vertex of the smaller base of the trapezoid, draw a line parallel to the diagonal.

## Solution

Through vertex $C$ of the smaller base $BC$ of trapezoid $ABCD$, draw a line parallel to diagonal $BD$ until it intersects line $AD$ at point $P$. Then

$$
AP = AD + DP = AD + BC = 2 \cdot 7 = 14
$$

Let $AC = x$, $BD = CP = y$. Since $\angle ACP = 120^{\circ}$, we have

$$
AC^2 + CP^2 - 2 \cdot AC \cdot CP \cdot \cos 120^{\circ} = AP^2, \text{ or } x^2 + y^2 + xy = 196
$$

On the other hand,

$$
S_{\triangle ACP} = \frac{1}{2} xy \cdot \sin 120^{\circ} = \frac{1}{2} AP \cdot \frac{15 \sqrt{3}}{7}
$$

or $xy = 60$. From the system

$$
\left\{\begin{array}{l}
x^2 + y^2 + xy = 196 \\
xy = 60
\end{array}\right.
$$

we find that $AC = x = 6$, $BD = y = 10$ or $AC = x = 10$, $BD = y = 6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-44.jpg?height=329&width=872&top_left_y=607&top_left_x=593)

Answer

6 and 10.

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