Elementary vectors question 
Relative to a fixed origin $O$, the points $A$ and $B$ have position
  vectors $\mathbf{a}$ and $\mathbf{b}$ repsectively, where $O,A$ and
  $B$ are not collinear. 
The point $C$ has position vector
  $\alpha\mathbf{a}+\beta\mathbf{b}$, where $\alpha,\beta$ are positive constants with
  $\alpha+\beta<1$. The lines $OA$ and $BC$ intersect
  at the point with position vector $\mathbf{x}$. 
Show that
  $$\mathbf{x}=\frac{\alpha}{1-\beta}\mathbf{a}.$$

My attempt:
The line segment $OA$ can be parametrized by $\mathbf{r}(t)=t\mathbf{a}, t\in[0,1]$, and the line segment $BC$ can be parametrized by $\mathbf{r}(s)=s(\alpha\mathbf{a}+(\beta-1)\mathbf{b})$. When $OA$ and $BC$ intersect,
$$t\mathbf{a}=s(\alpha\mathbf{a}+(\beta-1)\mathbf{b}).$$
It's not clear to me how to proceed from here. I can take the scalar product of both sides with $\mathbf{a}$ or $\mathbf{b}$, but this doesn't seem to help. 
A hint would be greatly appreciated.
 A: Actually, $\mathbf r(s)$ should be this:
$$\mathbf r(s)=\mathbf b+s((\alpha\mathbf a+\beta\mathbf b) - \mathbf b) = (1-s)\mathbf b +s(\alpha \mathbf a+\beta \mathbf b)$$
which, with $\mathbf r(t)=t\mathbf a$, gives you:
$$t\mathbf a=(1-s)\mathbf b +s(\alpha \mathbf a+\beta \mathbf b)$$
or, after tidying up:
$$(t-s\alpha)\mathbf a+(-(1-s)-s\beta)\mathbf b=0$$
Now, recall that $O,A,B$ are not colinear, so $\mathbf a$ and$\mathbf b$ are linearly independent, which implies:
$$\begin{align}t-s\alpha=0\\-(1-s)-s\beta=0\end{align}$$
which is a system of equations to solve for $s,t$. The second equation gives us $(1-\beta)s=1$, i.e. $s=\frac{1}{1-\beta}$. Substituting in the first, we get $t=s\alpha=(\frac{1}{1-\beta})\alpha=\frac{\alpha}{1-\beta}$, as desired.
A: HINT: We can draw a representative image as the following and use the similarity between triangles $XDC$ and $XOB$ in order to find the magnitude of $\vec{x}$ as $$|\vec{x}| =\frac{\alpha}{1-\beta} |\vec{a}|$$

Notice that we draw the representative image by considering the facts $\alpha+\beta<1$ (otherwise $\vec{c}$ would intersect the line $AB$ shown as gray) and $\alpha, \beta > 0$ (otherwise, $\vec{c}$ wouldn't lie inside the triangle $AOB$) so that the representative is valid.
