以知AM平行BN,AC平分角MAB,BC平分角NBA 1 过点C作直线DE,分别交AM,BN于点D,E ,求证:AB=AD+BE

问题描述:

以知AM平行BN,AC平分角MAB,BC平分角NBA 1 过点C作直线DE,分别交AM,BN于点D,E ,求证:AB=AD+BE

AB=AD+BE

你去菁优网上看看吧

延长AC交BN于点F.在△ABC和△FBC中,∠ACB = 90°= ∠FCB ,BC为公共边,∠ABC = ∠FBC ,所以,△ABC ≌ △FBC ,可得:AB = FB ,AC = FC ;则有:∠BFA = ∠BAF = ∠FAD .在△ACD和△FCE中,∠ACD = ∠FCE ,AC = FC ,∠CA...