Question

如图所示,三角形ABC的三条角平分线AD,BE,CF相交于点O,试说明角AOE和角1的关系（角1是角BCF）

Answer 1

∵OB,OA是∠ABC和∠BAC的角平分线

∴∠OAB+∠OBA = （∠ABC+∠BAC）/2 = (180°-∠ACB)/2

∵∠AOE = ∠OAB+∠OBA

∴∠AOE = 90°-∠ACB/2

∵OC是∠ACB的角平分线

∴∠FCB = ∠ACB/2

∴∠AOE = 90°-∠FCB

∴∠AOE和∠FCB互余