如图,三角形ABC中,角ACB=90度,CD垂直AB于D,AO平分角BAC,交CD于O,OE//BC交AB于E,求证：AC=AE

∵AO平分∠BAC∴∠CAO=∠BAO∵OE∥CB∴∠DCB=∠DOE又∵CD⊥AB∴∠ACB＝∠CDB＝90°∴∠ACD＝∠ACB－∠DCB　∠DEO＝90°－∠DOE∴∠ACD＝∠DEO在△ACO与△AEO中　→∠ACD=∠DEO　　∠CAO=∠BAO　　AO=AO　→△ACO≌△...