$E$ and $F$ are respectively the mid-points of the non-parallel sides $AD$ and $BC$ of a trapezium $ABCD$. Prove that $EF \parallel AB$ and $EF = \frac{1}{2}(AB + CD)$.

(N/A) Given: $A$ trapezium $ABCD$ in which $E$ and $F$ are respectively the mid-points of the non-parallel sides $AD$ and $BC$.
To prove: $EF \parallel AB$ and $EF = \frac{1}{2}(AB + CD)$.
Construction: Join $DF$ and produce it to intersect $AB$ produced at $G$.
Proof: In $\Delta CFD$ and $\Delta BFG$,we have:
$DC \parallel AG$ (since $DC \parallel AB$)
$\angle DCF = \angle GBF$ [Alternate interior angles]
$CF = BF$ [Given,$F$ is the mid-point of $BC$]
$\angle CFD = \angle BFG$ [Vertically opposite angles]
So,by $ASA$ criterion of congruence,$\Delta CFD \cong \Delta BFG$.
Therefore,$CD = BG$ and $DF = FG$ [$CPCT$].
Now,in $\Delta ADG$,$E$ is the mid-point of $AD$ and $F$ is the mid-point of $DG$ (since $DF = FG$).
By the Mid-point theorem,$EF \parallel AG$ and $EF = \frac{1}{2}AG$.
Since $AG = AB + BG$ and $BG = CD$,we have $AG = AB + CD$.
Thus,$EF \parallel AB$ and $EF = \frac{1}{2}(AB + CD)$.
Hence,proved.