已知：如图,AB‖CD,AM,CM分别平分∠BAC和角ACD,交与点M,过M作一直线交AB于E,交CD于F,求证：AC=AE+C

在AC上取点G,使AE＝AG.∵AE＝AG,AM＝AM,∠EAM＝∠GAM,∴△AEM≌△AGM,∴EM＝GM,∠AEM＝∠AGM.∵AE∥CF,∴∠CFM＝180°－∠AEM,显然有：∠CGM＝180°－∠AGM,又∠AEM＝∠AGM,∴∠CGM＝∠CFM,又CM＝CM,∠GCM＝∠FCM,△...