Question

已知△ABC中,AD平分∠BAC,EF⊥AD于点G,交AB于点E,交BC的延长线于点M,求证：∠M=（1/2∠ACB-∠

Answer 1

证明：∵EF⊥AD,AD平分∠BAC,

∴∠1=∠2,∠APE=∠APF=90°,

又∵∠AEF=180°-∠1-∠APE,∠AFE=180°-∠2-∠APF,

∴∠AEF=∠AFE,

∵∠CFM=∠AFE,

∴∠AEF=∠AFE=∠CFM,

∵∠AEF=∠B+∠M,∠MFC=∠ACB-∠M,

∴∠B+∠M=∠ACB-∠M,

∴∠M=1/2(∠ACB-∠B)