Find the ratio of AM:MO In a $∆ABC$, $BD$ and $CE$ are two medians which intersect each other at ‘$O$’. $AO$ intersects the line $ED$ at $M$. Find the ratio of $AM: MO$ $?$
 A: Doing it by the method of vectors would be easiest I think. Consider a triangle $ABC$.
Take point $B$ to be origin, $A$ to be pointed by a and $C$ by c,
now $O$ is centroid(point of intersection of medians) hence is pointed by $($a$+$b$+$c$)/3$ here, $O\equiv($a$+$c$)/3$.

So the vector corresponding to M can be calculated by considering the point $M$ divides $AO$ in ratio $k:1$ $(=AM:MO)$ and segment $ED$ in ratio $h:1$ $(=EM:MD)$. Calculating the vector corresponding to M using the two ratios gives two vectors(eg. finding $M$ from $h$:- $ M\equiv$ {( $h$$($a$+$c$)/2$ + a$/2$)}{$h+1$}which are equal as they point to the same point, and comparing the coefficients of a and c in them, gives the values of $h$ and $k$. Hence the answer $k$. 
A: $AO$ is the third median, and let it intersect $BC $ at $N $. Now, let us denote $AN=a $. We know $AO=\frac {2}{3}a $ ($O$ - barycenter) and $AM=\frac {1}{2}a $ (from the homothety with the center in $A $ and coefficient $\frac {1}{2}$). Thus, $MO=\frac {1}{6}a $ and $AM:MO=\frac {1}{2}a : \frac {1}{6}a = 3:1$.
