$\left[\begin{array}{l}\text { The Law of Sines } \\ {[\text { Transferring sides, diagonals, etc. }]}\end{array}\right]$

One of the lateral sides of the trapezoid forms an angle $\boldsymbol{\alpha}$ with the larger base, and the second side is equal to $a$ and forms an angle $\beta$ with the smaller base ( $\beta>\alpha$ ). Find the midline of the trapezoid if the smaller base is equal to $b$.
Through the vertex of the smaller base of the trapezoid, draw a line parallel to the opposite lateral side.

## Solution

Let $BC$ be the smaller base of trapezoid $ABCD$, where $\angle BAD = \alpha$, $\angle BCD = \beta$, $BC = b$, and $CD = a$. Through vertex $C$, draw a line parallel to the lateral side $AB$, intersecting the base $AD$ at point $K$. Then,

$$
\angle BCK = \angle BAD = \alpha, \angle DCK = \angle BCD - \angle BCK = \beta - \alpha
$$

By the Law of Sines in triangle $DCK$, we find that

$$
DK = \frac{CD \sin \angle DCK}{\sin \angle DKC} = \frac{a \sin (\beta - \alpha)}{\sin \alpha}
$$

Therefore, the midline of trapezoid $ABCD$ is

$$
\begin{gathered}
\frac{1}{2}(BC + AD) = \frac{1}{2}(BC + AK + KD) = \frac{1}{2}(2BC + KD) = \\
BC + \frac{1}{2} KD = b + \frac{a \sin (\beta - \alpha)}{2 \sin \alpha}
\end{gathered}
$$

## Answer

$b + \frac{a \sin (\beta - \alpha)}{2 \sin \alpha}$.