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# Help needed ASAP!

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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?

Hint(s):

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

Sep 23, 2018

#1
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$$Consider\;\; \triangle BDC\;\;and\;\;\triangle BPN\\ \angle PBN=\angle DBC \;\qquad common\;\;angle\\ \frac{\overline{ BD}}{\overline{BP}}=\frac{\overline{ BC}}{\overline{BN}}=2\\ \therefore \triangle BPN \sim \triangle BDC\\ \qquad \text{One equal angle and leg lengths from it in the same ratio.}\\ So \;\;\overline{DC}=2 \cdot \overline{PN} \qquad \text{ corresponding sides in similar triangles with a dilation factor of 2}$$

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

$$\overline{AB}=2\cdot \overline{MP}$$

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Sep 25, 2018
edited by Melody  Sep 25, 2018
#2
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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  $$AB \| DC$$

Sep 25, 2018
edited by Melody  Sep 25, 2018