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?


Let P be the midpoint of diagonal \(\overline{BD}\). What does the triangle inequality tell you about triangle M N P?

Guest Sep 23, 2018



\(Consider\;\; \triangle BDC\;\;and\;\;\triangle BPN\\ \angle PBN=\angle DBC \;\qquad common\;\;angle\\ \frac{\overline{ BD}}{\overline{BP}}=\frac{\overline{ BC}}{\overline{BN}}=2\\ \therefore \triangle BPN \sim \triangle BDC\\ \qquad      \text{One equal angle and leg lengths from it in the same ratio.}\\ So \;\;\overline{DC}=2 \cdot \overline{PN}      \qquad \text{ corresponding sides in similar triangles with a dilation factor of 2}\)




Using the same logic with \(\triangle APD\;\; and\;\; \triangle MPD\) it can be shown that  

\(\overline{AB}=2\cdot \overline{MP} \)

Melody  Sep 25, 2018
edited by Melody  Sep 25, 2018

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  \(AB \| DC\)


Melody  Sep 25, 2018
edited by Melody  Sep 25, 2018

11 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.