Question

如图,三角形ABC,角平分线AB、BE、cF相交于点H,过H,点作HG垂直于Ac,垂足为G,那么角AHE等于角cHG吗?

Answer 1

证明：

∵AD平分∠BAC

∴∠BAD＝∠BAC/2

∵BE平分∠ABC

∴∠ABE＝∠ABC/2

∴∠AHE＝∠BAD+∠ABE＝（∠BAC+∠ABC）/2＝（180-∠ACB）/2＝90-∠ACB/2

∵CF平分∠ACB

∴∠ACF＝∠ACB/2

∵HG⊥AC

∴∠CHG+∠ACF＝90

∴∠CHG＝90-∠ACF＝90-∠ACB/2

∴∠AHE＝∠CHG