题目内容
(2011?江汉区)已知?ABCD的周长为28,自顶点A作AE⊥DC于点E,AF⊥BC于点F.若AE=3,AF=4,则CE﹣CF= .
14﹣7
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或2﹣
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.
如图1:
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∵AE⊥DC,AF⊥BC,
∴∠AED=∠AFB=90°,
∵四边形ABCD是平行四边形,
∴∠ADC=∠CBA,AB=CD,AD=BC,
∴△ADE∽△ABF,
∴
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,
∵AD+CD+BC+AB=28,
即AD+AB=14,
∴AD=6,AB=8,
∴DE=3
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,BF=4
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,
∴EC=CD-DE=8-3
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,CF=BF-BC=4
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-6,
∴CE-CF=(8-3
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)-(4
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-6)=14-7
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;
如图2:∵AE⊥DC,AF⊥BC,
∴∠AED=∠AFB=90°,
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∵四边形ABCD是平行四边形,
∴∠ADC=∠CBA,AB=CD,AD=BC,
∴∠ADE=∠ABF,
∴△ADE∽△ABF,
∴
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,
∵AD+CD+BC+AB=28,
即AD+AB=14,
∴AD=6,AB=8,
∴DE=3
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,BF=4
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,
∴EC=CD+DE=8+3
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,CF=BC+BF=6+4
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,
∴CE-CF=(8+3
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)-(6+4
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)=2-
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.
∴CE-CF=14-7
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或2-
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.
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