CA = CO，

∴AD=DO，

In Rt△ACB, cos∠CAB=

1

three

=

Alternating current

ab blood type

=

six

ab blood type

∴AB=3AC= 18，

In Rt△ADC: cos∠CAB=

1

three

=

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Alternating current

∴AD=

1

three

AC=2，

∴AO=2AD=4，

∴BO=AB-AO= 18-4= 14，

∑△ AC ′ b ′ is obtained by △ △ACB rotation.

∴ac=ac′,ab=ab′,∠cac′=∠bab′，

∫∠ACC′=

1

2

( 180 -∠CAC ')，∠ABB ' =

1

2

( 180-∠BAB′)，

∴∠abb′=∠acc′，

∴, ∠BFO=∠CAO in △CAO and △BFO,

CA = CO，

∴∠COA=∠CAO，

∠∠COA =∠BOF (equal to the vertex angle),

∴∠BOF=∠BFO，

∴BF=BO= 14.

So the answer is: 14.