连接AF 和CE ∵AC EF 互相平分∴四边形AECF是平行四边形∴AB∥CD且AF=CE ∠AFC=∠AEC∴∠AFD=∠CEB易证△AFD≡△CEB∴∠B=∠D=90°∵AB∥CD∴∠D=∠DAB=90°=∠B∴四边形ABCD为矩形
