Question

AD平分角BAC,DG垂直于BC且平分BC,DE垂直于AB于E,DF垂直于AC于F.1.说明B

Answer 1

(1)证明:连接DB,DC.

DG垂直平分BC,则DB=DC;

DE垂直AB,DF垂直AC,AD平分角BAC,则DE=DF.

故Rt⊿DEB≌Rt⊿DFC(HL),得:BE=CF.

(2)解:DE=DF(已证);AD=AD.

则Rt⊿AED≌Rt⊿AFD(HL),AE=AF.

故AB+AC=(AE+BE)+(AF-CF)=AE+AF=2AE,即a+b=2AE,AE=(a+b)/2;

AB-AC=(AE+BE)-(AF-CF)=(AE+BE)-(AE-CF)=2BE,a-b=2BE,BE=(a-b)/2.