Vector trapezoid 
In the trapezoid $ABCD$ the sides $AB$ and $CD$ are parallel. In what ratio do the
  diagonals cut each other if $|AB|=3|CD|$ ?

Not too sure how to approach this question. So far I have attempted to find the intersection of the diagonals and called it point $M$.
Hence, I have gotten vector $\vec{AM} =\vec{AC} + t\cdot\vec{CB}$ where $t$ is a real value for example.
vector $\vec{AM}$ however can be rewritten as $s\cdot\vec{AD}$ where $s$ is a real value and this is equivalent to $s\cdot (\vec{AC} + \vec{CD})$
Not too sure what to do now, any ideas?? 
 A: Hint. Let $M$ be the intersection point of the two diagonals $AC$ and $DB$. Then, since $AB$ and $CD$ are parallel, it follows that the triangles $\triangle MCD$ and $\triangle MAB$ are similar.
A: Let the diagonals' intersection point be $X.$
Triangle $XAB$ is similar to triangle $XCD.$ So

 $$\frac{XA}{XC}=\frac{AB}{CD}=3\\ XA:XC:3:1$$ and $$\frac{XB}{XD}=\frac{AB}{CD}=3\\ XB:XD:3:1.$$


However, if a vector method is required:
Let the diagonals' intersection point be $X,$ and $$\vec{DC}=\mathbf{p},\\
\vec{DA}=\mathbf{q},\\
\vec{DX}=\lambda\,\vec{DB},\\
\vec{AX}=\mu\,\vec{AC}.$$
Now we form an equation relating $\lambda, \mu, \mathbf{p},$ and $\mathbf{q}:$

 $$\vec{DX}=\vec{DA}+\vec{AX}\\ \lambda\,\vec{DB}=\mathbf{q}+\mu\,\vec{AC}\\ \lambda(\mathbf{q}+3\mathbf{p})=\mathbf{q}+\mu(\mathbf{p}-\mathbf{q})\\ (3\lambda-\mu)\mathbf{p}+(\lambda+\mu-1)\mathbf{q}=\mathbf{0}.$$

Since $\mathbf{p}$ and $\mathbf{q}$ are not collinear, the coefficients of both vectors are $0.$

 Therefore, $$\lambda=\frac14 \\ DX=\frac14 DB \\ DX:XB=1:3$$ and $$\mu=\frac34 \\ AX=\frac34 AC \\ AX:XC=3:1.$$

