Help pleaseee

It doesn't specify what type of quadrilateral, so you can use any kind, and I would think a square would be the easiest to prove. Draw the diagram, label the points, and see if you can prove it from there. 

Hope this helps!

Let $ABCD$ be a convex quadrilateral, and let $M$ and $N$ be the midpoints of sides $\overline{AD}$ and $\overline{BC}$, respectively.
Prove that $MN \le (AB + CD)/2$.

When does equality occur?

Rotate the convex quadrilateral:

In the triangle  C'AB we must have $\overline{AB}+\overline{C'D'} < 2\overline{NM}$

Since $\overline{C'D'} = \overline{CD}$

$\begin{array}{|rcll|} \hline \overline{AB}+\overline{CD} < 2\overline{NM} \\\\ \dfrac{\overline{AB}+\overline{CD}}{2} < \overline{NM} \\ \hline \end{array}$

Equality occur if the convex quadrilateral is a trapezoid:

$\begin{array}{|rcll|} \hline \overline{AB}+\overline{CD} &=& 2 \overline{NM} \\\\ \dfrac{\overline{AB}+\overline{CD}}{2}& =& \overline{NM} \\ \hline \end{array}$