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.

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 $$ \vec r = t\vec c + \left(\frac {t(k-1)+1}2\right)\vec a\,. $$

(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.

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.