Question

四边形ABCD中ac平分角BAD,CE垂直于AB于E,AB+AD=2AE,求证角ADC+角B=180°,

Answer 1

证明：作CF⊥AD,交AD延长线于F∵CE⊥AB∴∠AFC=∠AEC=90º又∵∠EAC=∠FAC【AC平分∠BAD】AC=AC∴⊿AEC≌⊿AFC（AAS）∴AE=AF,CE=CF∵AB+AD=2AEAB=AE+BE∴AD+BE=AE=AF=AD+DF∴BE=DF又∵∠CFD=∠CEB=90º∴⊿CFD≌⊿CEB（SAS）∴∠B=∠FDC∵∠ADC+∠FDC=180º∴∠ADC+∠B=180º