(1)连接BD、CD
因DG垂直且平分BC,所以:BD=CD
AD平分∠BAC且DE⊥AB于E,DF⊥AC于F
DE=DF
∠DEB=∠DFC=90°
△BDE≌△CDF
BE=CF
(2)DE=DF,AD=AD,∠AED=∠AFD=90°
△AED≌△AFD,则:AE=AF
AB-BE=AC+CF=AC+BE
BE=(AB-AC)/2=(a-b)/2
AE=AB-BE=a-(a-b)/2=(a+b)/2
(1)连接BD、CD
因DG垂直且平分BC,所以:BD=CD
AD平分∠BAC且DE⊥AB于E,DF⊥AC于F
DE=DF
∠DEB=∠DFC=90°
△BDE≌△CDF
BE=CF
(2)DE=DF,AD=AD,∠AED=∠AFD=90°
△AED≌△AFD,则:AE=AF
AB-BE=AC+CF=AC+BE
BE=(AB-AC)/2=(a-b)/2
AE=AB-BE=a-(a-b)/2=(a+b)/2