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.

(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.
