# Proving that point of intersection of diagonals of trapezium lies on the line passing through the midpoint of parallel sides?

Can someone tell me the steps to solve this problem?

Prove by vector method that the point of intersection of diagonals of trapezium lies on the line passing through the midpoint of parallel sides.

I assumed the ratio of division of diagonals at intersection point as $a:1$ and $b:1$ and wrote the position vectors of the intersection point in terms of $a$ and $b$.But i'm getting too many variables.Any easier way out?

$$\vec{OM}=\frac 12 (\vec{OB}+\vec{OC}); \vec{ON}=\frac12(\vec{OA}+\vec{OD})$$ $\vec{OB}=k\vec{OA}$ and $\vec{OC}=k_1\vec{OD}$ $$\triangle BOC \sim \triangle AOD \Rightarrow \frac{OB}{OA}=\frac{OC}{OD}.$$ Then $k=k_1$.
Now $$\vec{OM}=\frac 12 (\vec{OB}+\vec{OC})=\frac12(k\vec{OA}+k\vec{OD})=$$ $$=k\frac12(\vec{OA}+\vec{OD})=k\vec{ON}$$ $\vec{OM}=k\vec{ON} \Rightarrow O,M,N -$ point on a straight line.