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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!

 Mar 21, 2019

it says a convex quadrilateral

Guest Mar 21, 2019

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:

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



 Mar 21, 2019
edited by heureka  Mar 21, 2019

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