Question

如图.已知AD平分∠BAC,CE⊥AD,交AB于E,EF平行于BCAC于F 求证：EC平分∠DEF

Answer 1

CE⊥AD且AD平分∠BAC,AD EC交于G 所以CG=GE,直角三角形CDG全等EDG,∠ECD=∠DEC；EF平行于BC,∠ECD=∠CEF,故,∠CEF=∠DEC,所以EC平分∠DEF.