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In triangle $\triangle ABC$, $AB=8$, $BC=14$ and $CA=10$. Points $M$, $N$, and $P $ are the midpoints of sides $AB$, $BC$, and $CA$, respectively. If $M$, $N$, and $P$ are connected to form a triangle, what is the perimeter of triangle $\triangle MNP$?

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Do you know proportionality? – Sigur Mar 3 '13 at 23:29
Have you tried drawing a picture? – JavaMan Mar 3 '13 at 23:30

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$\displaystyle \frac{8}{14}=\frac{4}{MP}\Rightarrow MP=7$

$\displaystyle \frac{14}{8}=\frac{7}{NP}\Rightarrow NP=4$

$\displaystyle\frac{14}{10}=\frac{7}{MN}\Rightarrow MN=5$


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It'd be easier if you prove that MN is parallel to AC and so on. Parallelogram properties after that. =) – hjpotter92 Mar 3 '13 at 23:50
Ah you are right. :P – Chris Mar 3 '13 at 23:51

It is well known that $NPM$ is similar to $ABC$ with the scale factor of $\frac{1}{2}$ (from $ABC$ to $NPM$). The perimiter of $NPM$ is therefore $\frac{8+14+10}{2}=16$.

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I think you need only the basic properties of middle segments (or middle lines):

Definition: The segment of line joining the middle points of two sides of a triangle is called middle segment.

Lemma: A middle segment of a triangle is parallel to the third side and its lengths is half the third side's.

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