Convex quadrilateral and triangle inequality

"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:

• 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. – achille hui Sep 22 '13 at 2:13
• 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? – Ozera Sep 22 '13 at 2:50
• 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. – achille hui Sep 22 '13 at 3:02
• 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 $\triangle MNP$. – achille hui Sep 22 '13 at 3:09
• 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 $\le 180^\circ$. – 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.

• Sorry, here is the image I made with P defined. i.imgur.com/QmRlpQF.png – Ozera Sep 22 '13 at 2:33
• @Ozera It's better to add that image to the question itself. – Calvin Lin Sep 22 '13 at 2:34
• Sorry, can you explain why AB || MN || CD ? and then why $P$ lies on $MN$? – Ozera Sep 22 '13 at 2:36
• That is the equality case. We have $AB \parallel MP = PN \parallel CD$, where the 'equality' is in terms of lines. – Calvin Lin Sep 22 '13 at 2:37
• 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 $<$? – Ozera Sep 22 '13 at 2:42