Question

已知,如图,在矩形(对边平行且相等,四个内角为直角)ABCD外一点E,联结AE,CE,CE交AD于F,角AEC=90度,

Answer 1

在ΔAFE与ΔCFD中,

∠AEF=∠D=90°,AF=CF,∠AFE=∠CFD,

∴ΔAFE≌ΔCFD(ASA),

∴AE=CD,∠EAF=∠DCF,

∵AB=CD,∴AB=AE,

∵AF=CF,∴∠FAC=∠FCA,

∴∠FAC+∠EAF=∠FCA+∠DCF,即∠EAC=∠DCA,

∵AB∥CD,∴∠BAC=∠DCA=∠EAC,

∴AC是等腰三角形ΔABE的顶角平分线,

∴AC⊥BE.