#1**+1 **

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!

LagTho Mar 21, 2019

#3**+2 **

**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:**

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

heureka Mar 21, 2019