Let AD be the angle bisector of the angle A of Delta ABC , then vec(AD) = alpha vec(AB) + beta vec(AC) , wh

### Question Asked by a Student from EXXAMM.com Team

Q 2221701621.     Let AD be the angle bisector of the angle A of Delta ABC , then

vec(AD) = alpha vec(AB) + beta vec(AC) , where
A

alpha = ( | vec(AB) | )/( | vec(AB) + vec(AC) |), beta = (|vec(AC) | ) /(|vec(AB) | + vec(AC) | )

B

alpha = ( | vec(AB) | + | vec(AC) )/( | vec(AB) |), beta = ( |vec(AB) | + |vec(AC) | ) /(| vec(AC) |)

C

alpha = ( | vec(AC) | )/( | vec(AB) + vec(AC) |), beta = (|vec(AB) | ) /(|vec(AB) | + vec(AC)|)

D

alpha = ( | vec(AB) | )/( | vec(AC) |), beta = (|vec(AC) | ) /(|vec(AB) | )

#### HINT

(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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