Let `vecAD` be the angle bisector of `/_A` of `Delta ABC` such that `vecAD=alphavecAB+betavecAC` THEN

Options

A. `alpha =(|AB|)/(|AB|+|AC|),beta =(|AC|)/(|AB|+|AC|)`
B. `alpha =(|AB|+|AC|)/(|AB|),beta =(|AB|+|AC|)/(|AC|)`
C. `alpha =(|AC|)/(|AB|+|AC|), beta =(|AB|)/(|AB|+|AC|)`
D. `alpha =(|AB|)/(|AC|), beta =(|AC|)/(|AB|)`

Answer

C

Explanation

Clearly ,AD divides BC in the ratio `AB:AC`.
`therefore AD =(|AB|AC||AC|AB)/(|AB|+|AC|)`
`implies |AD|= alpha AB + beta AC , ` where
` alpha =(|AC|)/(|AB|+|AC|) and beta =(|AB|)/(|AB|+|AC|)`