题目内容
如图,四边形ABCD中,DC∥AB,BC=1,AB=AC=AD=2,则BD的长为( ▲ )
A. | B. | C.3 | D.2 |
B
作AM⊥BC于点M,AN⊥BD于点N,
∵AC=AB,∴△ABC为等腰三角形,∴AM也是△ABC的中线(三线合一),
∴△ABM≌△ACM,∴∠CAM=∠BAM,
∵AB∥CD,AC=AD,∴∠ADC=∠ACD=∠CAB,
∵∠ADB=∠ABD=∠CDB,∴∠ADB=∠ADC=∠MAB,∴∠MAB=∠DBA,
又∵AB=AB,∴△ABN≌△BAM(AAS),∴AN=BC=,
∵AB=2,∴ =,∴ =15.∴BD=,故选B.
∵AC=AB,∴△ABC为等腰三角形,∴AM也是△ABC的中线(三线合一),
∴△ABM≌△ACM,∴∠CAM=∠BAM,
∵AB∥CD,AC=AD,∴∠ADC=∠ACD=∠CAB,
∵∠ADB=∠ABD=∠CDB,∴∠ADB=∠ADC=∠MAB,∴∠MAB=∠DBA,
又∵AB=AB,∴△ABN≌△BAM(AAS),∴AN=BC=,
∵AB=2,∴ =,∴ =15.∴BD=,故选B.
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