Question

已知△ABC中,角BAC=90°,AD⊥BS于D,CE平分∠ACD,交AD于G.交AB于E,EF⊥BC于F,证明

Answer 1

因为,∠AGE = ∠CGD = 90°-∠BCE = 90°-∠ACE = ∠AEG ,

所以,AE = AG .

因为,AE和EF是∠ACD平分线CE上一点E到∠ACD两边的距离,

所以,AE = EF .

因为,AG⊥BC ,EF⊥BC ,

可得：AG‖EF ,而且,AG = AE = EF ,

所以,四边形AEFG是平行四边形,

可得：AE = FG .

因为,AE = EF = FG = AG ,

所以,四边形AEFG是菱形.