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.

Solutions Collecting From Web of "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"

(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.

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