Question

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=

Answer 1

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,△CGM≌△CFM,∴CG＝CF.

由AE＝AG,CG＝CF,得：AG＋CG＝AE＋CF,即：AC＝AE＋CF.