How to prove that the line joinning the midpoints of the diagonals of trapezium is parallel to the parallel sides of the trapezium? 
How to prove that the line joining the midpoints of the diagonals of trapezium is parallel to the parallel sides of the trapezium?

I tried to prove this by drawing BD and joining the midpoints of AD and BC. I am unable to prove that this line and the line joining the midpoints of the diagonals are the same, please help
 A: Perhaps it is not surprising that all the segments (not only those that are actually depicted) going through the midpoint of the red line are halved by the same red point:

Taking any trapezium the two diagonals will be parallel to two of the colored lines above. That is, their midpoints will be equidistant to both of the black parallel lines...
A: let $ABCD$ be trapezium $M$ be mid points of diagonals $AC$&$P$ be mid points of $BC$
by mid point theorem 
$PM$ is parallel to$AB$
$PM$ intersect$BD$ at $N$
so $PN$ is parallel to $AB\implies $\parallel CD to$CD$ 
hence by converse of mid point theorem
$N$ is mid point of $BD$
again it meet $DA$ at $Q$ 
now $NQ$ is parallel to $CD$
by converse again q is mid point of $DA$
hence proved
A: Let's create an orthonormal basis that uses one of the two parallel sides of your trapezium as its first axis (let's call it x-axis).
Since the sides are parallel, every single point of the second side of the trapezium (that hasn't been used as a base axis) has its second coordinate equal to the same constant value C.
Now, the midpoints of the two diagonals have both their second coordinate equal to C/2 (this coordinate is (C+0)/2).
The straight line that goes through those two points is described by the equation :
$$
y = \frac{C}{2}
$$
and is parallel to the x-axis.
Since the x-axis is also parralel to the other side of our trapezium (by definition) the three straights are parallel to each other.
Sorry for my english I may have said a lot of faults.
