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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?
Mar 21, 2019

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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
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Guest Mar 21, 2019
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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?

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