Question

如图所示,在△ABC中,BC边的垂直平分线DE与∠BAC的平分线交于E,D为垂足,EF⊥AB交AB延长线于F,EG⊥AC

Answer 1

证明:(1)连接BE,CE.

DE垂直平分BC,则BE=CE;

EF垂直AB,EG垂直AC,AE平分角BAC,则EF=EG;

所以,Rt⊿EFB≌RtΔEGC(HL),得:BF=CG.

(2)EF=EG,AE=AE,则Rt⊿EFA≌RtΔEGA(HL),AF=AG.

故(1/2)(AB+AC)=(1/2)*[(AF-BF)+(AG+CG)]=(1/2)*(AF+AG)=(1/2)*(AF+AF)=AF.