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

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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


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

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