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选D. √3
解:
因为BC=√3BD,
|AD|=1AC.AD
= (AD+DC). AD
= |AD|² + DC.AD
= 1 + (BC - BD). AD
= 1 + (√3BD - BD ).AD
= 1 + (√3-1)BD.AD
= 1 + (√3-1)|BD| |AD| cos ∠BDA
=1 + (√3-1 |AD|×(BD cos ∠BDA)
= 1 + (√3-1) |AD|²
= √3
解:
因为BC=√3BD,
|AD|=1AC.AD
= (AD+DC). AD
= |AD|² + DC.AD
= 1 + (BC - BD). AD
= 1 + (√3BD - BD ).AD
= 1 + (√3-1)BD.AD
= 1 + (√3-1)|BD| |AD| cos ∠BDA
=1 + (√3-1 |AD|×(BD cos ∠BDA)
= 1 + (√3-1) |AD|²
= √3
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