Sum of coefficients of vectors equal to 0. In triangle ABC, D is the midpoint of side AB of triang1e ABC, and E is one·third of the way between C and B. Let F be intersection of AE and CD. Using vectors show that F is midpoint of CD.
I proved via showing that $F=\frac{C+D}{2}$ is on the line defined by A and E easily, is this correct? 
Also from book solution they easily showed that $t \vec{CF}-3\vec{CE}-\vec{CA}=0$ and since A,F,E are collinear $t-3-1=0$ but how?
Thanks 
 A: 
they easily showed that $t \vec{CF}-3\vec{CE}-\vec{CA}=0$ and since A,F,E are collinear $t-3-1=0$ but how?

Take $C$ to be the origin of coordinates, $C=0$.  From now on we use vectors based at $0$ (a vector space).
To say that the terminal point of vector $v$ is collinear with the termini of vectors $x$ and $y$ is to say that $v = ax + (1-a)y$ for some real value of $a$.  This is equivalent to having an equation $Av + Bx + Cy = 0$ with $A+B+C=0$, for example $A=1, B=-a, C=a-1$ and multiples of that.  As long as $v,x,y$ are nonzero there are no other linear equations relating them with $A+B+C \neq 0$. 
A: I don't think you need vectors for this problem. Just take $L$ to be the midpoint of segment $BE$. Then since $BE : EC = 2:1$ and $BL = LE$, we see that $BL = LE = EC$. Consequently, segment $DL$ is a mid-segment in triangle $ABE$ so $DL$ is parallel to $AE$. Since $F$ lies on $AE$, segment $EF$ is parallel to $DL$. However, $E$ is the midpoint of segment $CL$ and as already shown $EF$ is parallel to $DL$, so $EF$ is a mid-segment of triangle $CLD$ and hence point $F$ is a midpoint of $CD$.
