The median AD of the triangle ABC is bisected at E and BE meets AC at F. Find AF:FC.

Explanation

Taking A at the origin
Let P.V. of B and C be `vecb and vecc`, respectively.
P.V. of D is `(vecb+vecc)/(2)` and P.V. of E is `(vecb+vecc)/(4)`
Let `AF : FC = p : 1`.
Then position vector of F is `(pvecc)/(p+1)" "`(i)
Let `BF:EF = q : 1`.
The position vector of F is `(q((vecb+vecc))/(4)-vecb)/(q-1)" "`(ii)
Comparing P.V. of F in (i) and (ii), we have
`" "(pvecc)/(p+1)=(q((vecb+vecc))/(4)-vecb)/(q-1)`
Since vector `vecb and vecc` are independent, we have
`" "(p)/(p+1)=(q)/(4(q-1)) and (q-4)/(4(q-1))=0`
`rArr" "p=1//4 and q=4`
`rArr" "AF : FC = 1:2`