4. Given an isosceles trapezoid with base $A D$ such that $\overline{A B}=\overline{C D}=a, \overline{A C}=\overline{B D}=b$ and $\overline{B C}=c$. Let $M$ be an arbitrary point on the arc $B C$ of the circumscribed circle around the trapezoid $A B C D$. Express $\frac{\overline{B M}+\overline{M C}}{\overline{A M}+\overline{M D}}$ in terms of $a, b$ and $c$.
Solution. Let $R$ be the radius of the circumscribed circle around the trapezoid $ABCD$; let us denote

$$
\measuredangle BAC = \alpha, \quad \measuredangle ABC = \beta, \quad \measuredangle BAM = \gamma
$$

Using the sine theorem, we get

$$
\begin{aligned}
& \overline{BM} = 2R \sin \gamma \\
& \overline{MC} = 2R \sin (\alpha - \gamma) \\
& \overline{AM} = 2R \sin (\gamma + \pi - (\alpha + \beta)) = 2R \sin (\alpha + \beta - \gamma) \\
& \overline{MD} = 2R \sin (\alpha - \gamma + \pi - \alpha - \beta) = 2R \sin (\beta + \gamma)
\end{aligned}
$$

Using this, we get

![](https://cdn.mathpix.com/cropped/2024_06_05_47c254b2217fdab00c02g-7.jpg?height=442&width=447&top_left_y=1255&top_left_x=1033)

$$
\begin{gathered}
\frac{\overline{BM} + \overline{MC}}{\overline{AM} + \overline{MD}} = \frac{\sin \gamma + \sin (\alpha - \gamma)}{\sin (\alpha + \beta - \gamma) + \sin (\beta + \gamma)} = \frac{2 \sin \frac{\alpha + \gamma - \gamma}{2} \cos \frac{\gamma - \alpha + \gamma}{2}}{2 \sin \frac{\alpha + \beta - \gamma + \beta + \gamma}{2} \cos \frac{\alpha + \beta - \gamma - \beta - \gamma}{2}} \\
= \frac{2 \sin \frac{\alpha}{2} \cos \left(\gamma - \frac{\alpha}{2}\right)}{2 \sin \left(\beta + \frac{\alpha}{2}\right) \cos \left(\gamma - \frac{\alpha}{2}\right)} = \frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \sin \left(\beta + \frac{\alpha}{2}\right) \cos \frac{\alpha}{2}} = \frac{\sin \alpha}{\sin (\alpha + \beta) + \sin \beta} \\
= \frac{2R \sin \alpha}{2R \sin [\pi - (\alpha + \beta)] + 2R \sin \beta} = \frac{c}{a + b}
\end{gathered}
$$