OG⊥BC,点O是△ABC三条角平分线的交点,
∠2=90°-∠BCE=90°-1/2∠启缓ACB,
∠1=∠DAB+∠DBA
=1/2(∠BAC+∠ABC)
=1/2(∠BAC+∠ABC+∠ACB-∠ACB)(三角形内角和180°)世宴
=90-1/2∠ACB=∠搜旁银2
因为OG⊥BC,点O是△ABC三条角枝稿平分高亩线的交点
所以∠2=90°-∠BCE=90°-1/2∠ACB
所以∠1=∠DAB+∠DBA
=1/2(∠BAC+∠ABC)
=1/猛念孝2(∠BAC+∠ABC+∠ACB-∠ACB)(三角形内角和)
=90-1/2∠ACB
=∠2