39. A trapezoid $A B C D$ is inscribed in a circle, with $[B C] \|[A D]$ and $|B C|<|A D|$. A diameter $[C M]$ is drawn through the center of the circle $O$. The line $B M$ intersects the side $[A D]$ at point $K$. Point $L$ is the projection of point $K$ onto the diameter $[C M]$. Calculate $|E F|$, if it is known that $|L C|=\frac{9}{8} R$, where $R$ is the radius of the circle, point $F$ is the midpoint of $[O M]$, and point $E$ is the midpoint of the diagonal $[B D]$.
39. Let's connect points $M$ and $D$ (Fig. 59) and denote the angles of triangle $MBD$ as $\alpha$, $\beta$, and $\gamma$. Let $|OL| = m; |OL| = |LC| - |OC|$. In triangle $EOF$, we have: $|OE| = R \cos \alpha, |OF| = \frac{R}{2}, \widehat{EOF} = \alpha + 2\gamma$. From this, we get:

$$
\begin{gathered}
|EF|^2 = R^2 \cos^2 \alpha + \frac{R^2}{4} - R^2 \cos \alpha \cos (\alpha + 2\gamma). \\
\alpha + 2\gamma = \alpha + \beta + \gamma + \gamma - \beta = \pi + \gamma - \beta, \text{ therefore } \\
\cos (\alpha + 2\gamma) = -\cos (\gamma - \beta)
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_5cdadc99c18e78f1502ag-106.jpg?height=659&width=643&top_left_y=1732&top_left_x=1069)

Fig. 59
and, therefore,

$$
\begin{gathered}
|EF|^2 = R^2 \cos^2 \alpha + \frac{R^2}{4} + R^2 \cos \alpha \cos (\gamma - \beta) = \\
= R^2 - R^2 \sin^2 \alpha + \frac{R^2}{4} + R^2 \cos \alpha \cos (\gamma - \beta) = \\
= \frac{5R^2}{4} - R^2 \left( \sin^2 \alpha - \cos \alpha \cos (\gamma - \beta) \right) = \\
= \frac{5R^2}{4} - R^2 \left( 1 - \cos^2 \alpha - \cos \alpha \cos \beta \cos \gamma - \cos \alpha \sin \beta \sin \gamma \right) = \\
= \frac{5R^2}{4} - R^2 \left( 1 - \cos \alpha (\cos \alpha + \cos \beta \cos \gamma) - \cos \alpha \sin \beta \sin \gamma \right)
\end{gathered}
$$

But $\cos \alpha = \cos (\pi - (\beta + \gamma)) = -\cos (\beta + \gamma) = -\cos \beta \cos \gamma + \sin \beta \sin \gamma$. Therefore,

$$
|EF|^2 = \frac{5R^2}{4} - R^2 (1 - 2 \cos \alpha \sin \beta \sin \gamma)
$$

Notice now that $|MK| = |MD| \cos \alpha, |MD| = 2R \sin \beta$, so $|MK| = 2R \cos \alpha \sin \beta$. Since $\widehat{MKN} = \gamma$, then $|ML| = |MK| \sin \gamma = 2R \cos \alpha \sin \beta \sin \gamma$ and, therefore, $m = R - |ML| = R(1 - 2 \cos \alpha \sin \beta \sin \gamma)$. Thus, $|EF|^2 = \frac{5}{4} R^2 - Rm$. Since $m = \frac{9}{8} R - R = \frac{1}{8} R$, we finally get:

$$
|EF|^2 = \frac{5}{4} R^2 - \frac{1}{8} R^2 = \frac{9}{8} R^2 ; |EF| = \frac{3 \sqrt{2}}{4} R
$$