Решение:
В треугольнике DKE, ∠DKE = 90°.
- \( \angle D + \angle 1 = 90° \).
- В треугольнике EKF, ∠EKF = 90°.
- \( \angle F + \angle 2 = 90° \).
- \( \angle 2 = 90° - \angle F \).
- \( \angle 1 = 90° - \angle D \).
- \( (90° - \angle F) - (90° - \angle D) = 40° \).
- \( 90° - \angle F - 90° + \angle D = 40° \).
- \( \angle D - \angle F = 40° \).
- \( \angle D = 40° + \angle F \).
- В треугольнике DEF: \( \angle D + \angle DEF + \angle F = 180° \).
- \( \angle DEF = \angle 1 + \angle 2 \).
- \( (40° + \angle F) + (\angle 1 + \angle 2) + \angle F = 180° \).
- \( 40° + 2\angle F + \angle 1 + \angle 2 = 180° \).
- \( 2\angle F + 40° + 40° = 180° \) (так как \( \angle 2 - \angle 1 = 40° \) и \( \angle 1 + \angle 2 = 90° \), то \( 2\angle 2 = 130° \), \( \angle 2 = 65° \) и \( \angle 1 = 25° \)).
- \( 2\angle F + 80° = 180° \).
- \( 2\angle F = 100° \).
- \( \angle F = 50° \).
- \( \angle D = 40° + 50° = 90° \).
- \( \angle 1 = 90° - \angle D = 90° - 90° = 0° \). Это невозможно.
- Пересмотрим \( \angle 1 + \angle 2 = 90° \).
- \( \angle 2 = \angle 1 + 40° \).
- \( \angle 1 + (\angle 1 + 40°) = 90° \).
- \( 2\angle 1 + 40° = 90° \).
- \( 2\angle 1 = 50° \).
- \( \angle 1 = 25° \).
- \( \angle 2 = 25° + 40° = 65° \).
- \( \angle D = 90° - \angle 1 = 90° - 25° = 65° \).
- \( \angle F = 90° - \angle 2 = 90° - 65° = 25° \).
Ответ: ∠1 = 25°, ∠2 = 65°, ∠D = 65°, ∠F = 25°.