Question

如图，已知AB、CD、EF相交于点O，EF⊥AB，OG为∠COF的平分线，OH为∠DOG的平分线，若∠AOC：∠COG=4：7，

求∠DOF、∠DOH的大小

Answer 1

解：
∵EF⊥AB
∴∠AOF＝∠BOF＝90
∵∠AOC/∠COG＝4/7
∴∠COG＝7∠AOC/4
∵OG为∠COF的平分线
∴∠COG＝∠FOG＝∠COG/2
∴∠COF＝2∠COG
∴∠COF＝7∠AOC/2
∵∠AOC+∠COF＝∠AOF
∴∠AOC+∠COF＝90
∴∠AOC+7∠AOC/2＝90
∴∠AOC＝20
∴∠COF＝∠AOF-∠AOC＝90-20＝70
∴∠COG＝∠FOG＝∠COF/2＝70/2＝35
∵AB、CD相交于点O
∴∠BOD＝∠AOC＝20
∴∠DOG＝∠BOD+∠BOF +∠FOG＝20+90+35＝145
∵OH平分∠DOG
∴∠DOH＝∠DOG/2＝145/2＝72.5

Answer 2

∠AOC = 4[ 90°÷ ( 4+7+7)] = 20°
∠DOB = ∠AOC = 20°
∴ ∠DOF = ∠DOB + 90°= 110°
∠GOF = 7[ 90°÷ ( 4+7+7)] = 35°
∴ ∠DOG = ∠DOF + 35°= 145°
∴ ∠DOH = ∠DOG ÷ 2 = 72.5°