(2004•襄陽)在△ABC中,BE平分∠ABC交AC于點E,ED∥CB交AB于點D,已知:AD=1,DE=2,則BC的長為( )
A.3
B.4
C.5
D.6
【答案】
分析:因為ED∥CB,所以∠1=∠3,又因為BE平分∠ABC,所以∠1=∠2.故∠2=∠3,根據(jù)等角對等邊,BD=DE=2,根據(jù)平行線分線段成比例定理,
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=
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,即
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=
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,解得BC=6.
解答:
解:∵ED∥CB,
∴∠1=∠3,
又∵BE平分∠ABC,
∴∠1=∠2,
∴∠2=∠3,
∴BD=DE=2,
又∵ED∥CB,
∴
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=
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,
∵AD=1,DE=2,
∴AB=AD+BD=AD+DE=3,
即
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=
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,
∴BC=6.
故選D.
點評:此題結(jié)合了平行線的性質(zhì)和平行線分線段成比例定理,構(gòu)思巧妙,是一道很好的題.