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Help! A math problem about the nature of angle bisector in the second day of junior high school! !
Very simple

∫OE∠ BOC ∴∠COE=∠BOE.

* Equal share ∠AOC ∴∠AOF=∠COF

∴∠COF=∠EOF+∠BOE

∠EOF=(90 -∠AOF)+∠BOE

=(90-∞COF)+∞BOE

=(90 -∠EOF-∠BOE)+∠BOE

=90 -∠EOF

∴2∠EOF=90

∴∠EOF=45