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?
Hint(s):
Let P be the midpoint of diagonal \(\overline{BD}\). What does the triangle inequality tell you about triangle M N P?
+118608 
\(Consider\;\; \triangle BDC\;\;and\;\;\triangle BPN\\ \angle PBN=\angle DBC \;\qquad common\;\;angle\\ \frac{\overline{ BD}}{\overline{BP}}=\frac{\overline{ BC}}{\overline{BN}}=2\\ \therefore \triangle BPN \sim \triangle BDC\\ \qquad \text{One equal angle and leg lengths from it in the same ratio.}\\ So \;\;\overline{DC}=2 \cdot \overline{PN} \qquad \text{ corresponding sides in similar triangles with a dilation factor of 2}\)
Using the same logic with \(\triangle APD\;\; and\;\; \triangle MPD\) it can be shown that
\(\overline{AB}=2\cdot \overline{MP} \)
+118608 I have to continue on a second post because it is not displaying properly.
\( \text{Now consider the points MPN, these points either form a triangle or they are collinear}\\ \text{If they form a line then MP+PN=MN}\\ \text{If they from a triangle then use this fact:}\\ \text{In any triangle the sum of any 2 side lengths must be longer than the third side.}\\ So\;\; \\ {MP}+{PN} \ge{MN}\\ 2 {MP}+2{PN}\ge2{MN}\\ AB+CD\ge2MN\\ 2MN\le AB+CD\\ MN \le \frac{AB+CD}{2} \)
There can only be an equality if P lies on the line MN in which case there is no triagle MPN and the the quadrilateral ABCD must be a trapezium (Australian definition, I think Americans call it a trapezoid) where \(AB \| DC\)
