Question

在△ABC中,AD平分∠BAC,EF⊥AD交AB、AC于E、F,交BC延长线于M,求证∠M=1/2(∠ACB－∠B）.

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,

即：∠M= 1/2（∠ACB-∠B）．

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