Prove that the point of intersection of the diagonals of the trapezium lies on the line passing through the mid-points of the parallel sides Prove,by vector method,that the point of intersection of the diagonals of the trapezium lies on the line passing through the mid-points of the parallel sides.

My Attempt:
Let the trapezium be $OABC$ and that the O is a origin and the position vectors of $A,B,C$ be $\vec{a},\vec{b},\vec{c}$.Then the equation of $OB$ diagonal is $\vec{r}=\vec{0}+\lambda \vec{b}................(1)$
And equation of $AC$ diagonal is $\vec{r}=\vec{a}+\mu(\vec{c}-\vec{a}).......(2)$
And the equation of the line joining the mid points of $OA$ i.e.$\frac{\vec{a}}{2}$ and $BC$ i.e. $\frac{\vec{b}+\vec{c}}{2}$ is $\vec{r}=\frac{\vec{a}}{2}+t(\frac{\vec{b}+\vec{c}}{2}-\frac{\vec{a}}{2}).........(3)$.
Here $\lambda,\mu,t$ are scalars.
I do not know how to solve $(1)$ and $(2)$ and put into $(3)$ to prove the desired result.
Please help me.Thanks.
 A: You're missing an additional piece of information: There are two parallel sides to the trapezium. Express this by assuming that $\vec b -\vec c = k\vec a$ for some $k$. You now can solve for 
$$\vec b = \vec c + k\vec a\,.\tag4$$
 Put (4) into the equation (1) = (2) to find
$$\mu = \lambda = \frac1{k+1}\,.$$
You then check that with this choice of $\lambda$ and $\mu$, the resulting $\vec r$ satisfies (3). Note that with (4), equation (3) becomes
$$
\vec r = t\vec c + \left(\frac {t(k-1)+1}2\right)\vec a\,.
$$
A: 
The trapezium's diagonals intersect at $X.$ We want to show that $B,X,\text{ and }R$ are collinear.$$ $$
Triangle $XOP$ is similar to triangle $XQA.$ So $$\frac{XO}{XQ}=\frac{OP}{QA}\\=\frac1\lambda.$$
Now, $$\vec{BX}=\vec{OX}-\vec{OB}\\=\frac1{\lambda+1}\vec{OQ}-\vec{OB}\\=\frac1{\lambda+1}\left(\mathbf{a}+2\lambda\mathbf{b}\right)-\mathbf{b}\\=\frac1{\lambda+1}\left[\mathbf{a}+\left(\lambda-1\right)\mathbf{b}\right],$$ while $$\vec{BR}=\vec{BO}+\vec{OA}+\vec{AR}\\=-\mathbf{b}+\mathbf{a}+\lambda\mathbf{b}\\=\mathbf{a}+\left(\lambda-1\right)\mathbf{b}.$$
Therefore $$\vec{BR}=(\lambda+1)\vec{BX}.$$ Thus $B,X,\text{ and }R$ are collinear.
A: (1) and (2) can be solved
for $\mu$ and $\lambda$
by setting
$\lambda \vec{b}
=\vec{a}+\mu(\vec{c}-\vec{a})
$
and looking at the
two coordinates.
This gives two equations
in two unknowns,
and so can be solved.
Putting in the values for
$\lambda \vec{b}
$
gives the coordinates of the point
of intersection.
Similarly, 
equation (3) gives
parameterizations for
the coordinates
of the line.
Looking at one of these,
say $x$,
you can solves for
the value of $t$
that gives that $x$ value.
That same value of $t$
should give the
$y$ coordinate of the
intersection.
I feel that there should be
a more purely vector approach,
but one does not occur to me
right now.
