解:(1)设∠COF=α,则∠EOF=90°﹣α, ∵OF是∠AOE平分线, ∴∠AOF=90°﹣α, ∴∠AOC=(90°﹣α)﹣α=90°﹣2α, ∠BOE=180°﹣∠COE﹣∠AOC,=180°﹣90°﹣(90°﹣2α), =2α,即∠BOE=2∠COF; (2)成立,设∠AOC=β,则∠AOF=, ∴∠COF=45°+=(90°+β), ∠BOE=180°﹣∠AOE, =180°﹣(90°﹣β), =90°+β, ∴∠BOE=2∠COF (3)∠DOE=180°﹣∠BOD﹣∠AOE,=180°﹣(60﹣)°﹣(90°﹣n°),=(30+n)° |