矩形ABCD中,AC、BD相交于点O,E为矩形ABCD外一点,若AE⊥CE,求证:BE⊥DE.
矩形ABCD中,AC、BD相交于点O,E为矩形ABCD外一点,若AE⊥CE,求证:BE⊥DE.
数学人气:657 ℃时间:2019-11-23 07:14:51
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证明:连接OE,在△AEC中,∵四边形ABCD是矩形,∴AC=BD,OC=OD,OA=OC,∴OA=OB=OC=OD,∵AE⊥EC,∴OE=OA.∴OE=OB=OD,∴∠OBE=∠OEB,∠OED=∠ODE.∵∠ODE+∠OED+∠OBE+∠OEB=180°,∴2(∠OEB+∠OED)=180°...
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