证明：连接OE∵矩形ABCD∴AC＝BD,OA＝OB＝OC＝OD∵AE⊥CE∴OE＝AC/2 （直角三角形中线特性）∴OE＝BD/2∴OE＝OB＝OD∴∠OBE＝∠OEB,∠ODE＝∠ODE∵∠OBE+∠ODE+∠BED＝180,∠BED＝∠OEB+∠OED∴2∠OEB+2∠OED＝180∴...