"Let $ABCD$ be a convex quadrilateral, and let $M$ and $N$ be the midpoints of sides $\overline{AD}$ and $\overline{BC}$, respectively. Prove that $MN$ <= $\frac{AB + CD}{2}$

My question is if $MN \leq MP + NP$ or if $MN < MP + NP$ by triangle $MNP$ inequality on triangle. The solution manual says that it is $MN \leq MP + NP$, but doesn't the triangle inequality state that $MN < MP + NP$ ?

Image of the convex quadrilateral: Convex Quadrilateral

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    $\begingroup$ The triangle inequality is $MN \le MP + NP$ for any three general points. It is a constraint imposed on the distance function for any metric spaces. The inequality $MN < MP + NP$ is a different thing. It is a necessary condition for 3 points form a "non-degenerate" triangle in Euclidean geometry. $\endgroup$ – achille hui Sep 22 '13 at 2:13
  • $\begingroup$ Sorry I don't understand. $MN \leq MP + NP$ is for any three general points? How does what I have differ from a normal triangle? $\endgroup$ – Ozera Sep 22 '13 at 2:50
  • $\begingroup$ If $MN = MP+NP$, then the 3 points $M, N, P$ are collinear with $P$ lies between $M$ and $N$. The corresponding triangle "degenerate" into a "line segment". In your case, this will happen when your $ABCD$ is a trapezoid. Look at Calvin's answer again and think what happens when $ABCD$ is a trapezoid. $\endgroup$ – achille hui Sep 22 '13 at 3:02
  • $\begingroup$ From the picture, I see you are using Geogebra. Inside Geogebra, drag your point $B$ to a position such that $AB \parallel CD$ and observe what happens to the $\triangle MNP$. $\endgroup$ – achille hui Sep 22 '13 at 3:09
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    $\begingroup$ Nope, a convex quadrilateral isn't defined like what you draw. In your picture, if you move $B$ vertically downward, your quadrilateral remains convex until it moves below the line $AC$. A geometric figure is convex iff it contains all line segments joining any two point of it. For planar polygons, this condition is equivalent to all internal angles are $\le 180^\circ$. $\endgroup$ – achille hui Sep 22 '13 at 3:45

The slight issue which you haven't addressed, is the position of point $P$. Here is the proof, and an explanation of why $P$ is important.

Let $P$ be the midpoint of diagonal $BD$. Then, we see that $MP$ is parallel to $AB$, and half of it. $NP$ is parallel to $CD$ and half of it.

Applying the triangle inequality to the points $M, N, P$, we get that

$$ MN \leq NP + PM = \frac{AB+CD}{2}.$$

Note that equality holds because it is possible for $MNP$ to be a straight line.

Under this scenario, we have $AB \parallel MN \parallel CD$, which is the case of a trapezoid. Conversely, it is clear to see that if $ABCD$ is a trapezoid, then $AB \parallel MN \parallel CD$ and hence $P$ lies on $MN$ so $MN = NP + PM$ and we do indeed have equality.

  • $\begingroup$ Sorry, here is the image I made with P defined. i.imgur.com/QmRlpQF.png $\endgroup$ – Ozera Sep 22 '13 at 2:33
  • $\begingroup$ @Ozera It's better to add that image to the question itself. $\endgroup$ – Calvin Lin Sep 22 '13 at 2:34
  • $\begingroup$ Sorry, can you explain why AB || MN || CD ? and then why $P$ lies on $MN$? $\endgroup$ – Ozera Sep 22 '13 at 2:36
  • $\begingroup$ That is the equality case. We have $AB \parallel MP = PN \parallel CD$, where the 'equality' is in terms of lines. $\endgroup$ – Calvin Lin Sep 22 '13 at 2:37
  • $\begingroup$ Mmm..I think i'm misunderstanding. I don't see why you said "applying triangle inequality..." and then we get an inequality with a $\leq$ and not $<$. Isn't the triangle inequality with $<$? $\endgroup$ – Ozera Sep 22 '13 at 2:42

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