Prove that the line segment joining the mid-points of the non-parallel sides of a trapezium is parallel to the parallel sides of the trapezium.

(N/A) Given: $A$ trapezium $ABCD$ in which $AB \parallel DC$,and $E$ and $F$ are the mid-points of the non-parallel sides $AD$ and $BC$ respectively.
To prove: $EF \parallel AB$ and $EF \parallel DC$.
Construction: Join $DF$ and produce it to intersect the line $AB$ produced at point $G$.
Proof: In $\Delta DCF$ and $\Delta GBF$,we have:
$\angle 1 = \angle 2$ [Alternate interior angles,as $DC \parallel BG$]
$\angle 3 = \angle 4$ [Vertically opposite angles]
$CF = BF$ [Given,as $F$ is the mid-point of $BC$]
By $AAS$ congruence criterion,$\Delta DCF \cong \Delta GBF$.
Therefore,$DF = GF$ and $DC = GB$ [by $CPCT$].
In $\Delta DAG$,$E$ is the mid-point of $AD$ and $F$ is the mid-point of $DG$ (since $DF = GF$).
By the Mid-point Theorem,the line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Therefore,$EF \parallel AG$.
Since $AG$ is the same line as $AB$,we have $EF \parallel AB$.
Also,since $AB \parallel DC$ and $EF \parallel AB$,it follows that $EF \parallel DC$.
Hence,$EF \parallel AB \parallel DC$.