Question

在△ABC中,AD⊥BC,AB平分∠BAC,AG⊥AE,CG是△ABC外角∠ACF若∠G-∠DAE=60°,则∠ACB=

Answer 1

⑴∠DAE=∠EAC-∠DAC=∠BAC -∠DAC=(180°-∠B-∠ACB)-(90°-∠ACB )

=90°-∠B-∠ACB-90°+∠ACB=∠ACB-∠B

⑵∠CAG=90°-∠BAC ∠G=∠DAE+60° ∠GCA=（180°-∠ACB）／2=90°-∠ACB

∠CAG+∠G+∠GCA=180°故有

∠B+∠DAE=30°

由⑴代入⑵得∠B+∠ACB-∠B=30° 得∠ACB=60°望采纳