如图,四边形ABCD中,AC平分∠BAD,CE⊥AB于E,AD+AB=2AE.求证:∠B+∠ADC=180°.

问题描述:

如图,四边形ABCD中,AC平分∠BAD,CE⊥AB于E,AD+AB=2AE.求证:∠B+∠ADC=180°.

证明:过C作CF垂直AD于F,∵AC平分∠BAD,∴∠FAC=∠EAC,∵CE⊥AB,CF⊥AD,∴∠DFC=∠CEB=90°,∴△AFC≌△AEC,∴AF=AE,CF=CE,∵2AE=AB+AD,又∵AD=AF-DF,AB=AE+BE,AF=AE,∴2AE=AE+BE+AE-DF,∴BE=DF,∵∠...