the median AB of `Delta ABC` is bisected at Eand BE is produced to meet the side AC in f.then the ratio `AF:FC=`

Options

A. `3//4`
B. `1//3`
C. `1//2`
D. `1//4`

Explanation

Let position vector of A w.r.t. B is a and that of C w.r.t. B is c.

Position vector fo D w.r.t.
`B=(0+c)/(2)=(c)/(2)`
Position vector of
`E=(a+(c)/(2))/(2)=(a)/(2)+(c)/(4)` . . . (i)
Let `AF:FC=lamda:1 and BE:EF=mu:1`
Position vector of `F=(lamdac+a)/(1+lamda)`
Now, position vector of
`E=(mu((lamdac+a)/(1+lamda))+1*0)/(mu+1)` . . . (ii)
From eqs. (i) and (ii), we get
`(a)/(2)+(c)/(4)=(mu)/((1+lamda)(1+mu))a+(lamdau)/((1+lamda)(1+mu))c`
`implies(1)/(2)=(mu)/((1+lamda)(1+mu))`
and `(1)/(4)=(lamdamu)/((1+lamda)(1+mu))`
`implieslamda=(1)/(2)`
`therefore (AF)/(AC)=(AF)/(AF+FC)=(lamda)/(1+lamda)=((1)/(2))/((3)/(2))=(1)/(3)`.