5. (20 points) Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $E$, and the rays $D A$ and $C B$ intersect at point $F$. The ray $B A$ intersects the circumcircle of triangle $D E F$ at point $L$, and the ray $B C$ intersects the same circle at point $K$. The length of segment $L K$ is $5$, and $\angle E B C=15^{0}$. Find the radius of the circumcircle of triangle $E F K$.

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# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c1ab4103bc5f9777fe4g-04.jpg?height=662&width=714&top_left_y=1214&top_left_x=725)

$\angle F L E = \angle F D E = \angle F K E = \alpha$, since these angles subtend the arc $F E$.

$\angle E B K = \angle F D E = \alpha$, since quadrilateral $A B C D$ is cyclic.

$\triangle B L F$ is isosceles, as $\angle F L B = \angle F B L = \angle E B K = \alpha$.

Then $\angle B F L = 180 - 2 \alpha \Rightarrow \sin \angle B F L = \sin 2 \alpha = \sin 30^{\circ} = \frac{1}{2} . \quad 2 R = \frac{L K}{\sin \angle L F K} = 10 \Rightarrow R = 5$.

Answer: 5.

| Criterion | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 20 |
| It is proven that $\triangle B L F$ is isosceles. | 15 |
| It is proven that $\angle E B K = \angle F D E = \alpha$. | 10 |
| It is proven that $\angle F L E = \angle F D E = \angle F K E = \alpha$. | 5 |
| The solution does not meet any of the criteria listed above. | 0 |
| Maximum score | 20 |