Use the provided image to determine the subject and grade level, and then output the answer in the specified JSON format.

Solution:

• The problem is from geometry, likely for middle school or early high school.
• We are given a circle with center O. Points A, B, C, and D are on the circle.
• Segments EA and EB are drawn, with perpendiculars from E to OA and OB at points F and L respectively.
• We are given that ∠LF = 90° and EO = OF.
• We need to prove that EO = OF.

Proof:

1. In triangle △EOF, we are given that ∠EFO = 90° and EO = OF. This is a contradiction. The problem statement likely meant to state that ∠EFO = 90° and EF = EL or some other condition that would lead to proving EO = OF.
2. However, if we assume there is a typo and the intention was to prove EO = OF based on other given information in a standard geometry problem context:
3. Consider triangles △EOF and △EOL.
4. We are given that ∠EFO = 90° and ∠ELO = 90° (implied by the diagram with right angle symbols).
5. Segment EO is common to both triangles.
6. If we were given that EF = EL, then by the Hypotenuse-Leg (HL) theorem for right triangles, △EOF ≅ △EOL.
7. This would imply that OF = OL, but not necessarily EO = OF.
8. Let's re-examine the given information: LE=LF=90°. This is also unusual notation. It is highly probable that ∠EFO = 90° and ∠ELO = 90°.
9. The statement LE=LF=90° might mean that the angles at F and L are 90 degrees.
10. The condition EO=OF is what we are asked to prove (Доказать). Therefore, this cannot be a given condition.
11. There seems to be a significant misunderstanding or typo in the provided problem statement and diagram. The statement EO=OF is listed under "Дано" (Given) but also under "Доказать" (To Prove). This is contradictory.
12. Assuming that the intention was: Given: ∠EFO = 90°, ∠ELO = 90°, and EF = EL. Prove: EO = OF.
13. Even with these assumptions, proving EO = OF is not directly possible without more information or a different geometric configuration. If EF = EL and ∠EFO = ∠ELO = 90°, then △EOF ≅ △EOL by LL congruence, leading to OF = OL and ∠EOF = ∠EOL.
14. If the diagram is correct and O is the center of the circle, and F and L are points on OA and OB respectively, and EF is perpendicular to OA, and EL is perpendicular to OB, and if EF=EL, then O must lie on the angle bisector of ∠AOB. In this case, OF would be the distance from O to OA, and OL would be the distance from O to OB.
15. Let's consider the possibility that the diagram implies OF and OL are chords or radii. However, F and L appear to be points on the radii OA and OB.
16. Given the direct contradiction (EO=OF is in both 'Given' and 'To Prove'), it is impossible to provide a mathematically sound proof.
17. If we ignore the contradiction and assume the goal is to prove EO=OF, and that ∠EFO = 90°, then for EO=OF to hold, △EOF must be an isosceles right triangle where EF = OF. This implies ∠EOF = 45°.
18. Without any other valid conditions, the problem as stated is unsolvable.

Conclusion: The problem statement contains a contradiction, as EO=OF is listed under both "Given" and "To Prove". Therefore, a valid proof cannot be constructed based on the provided information.