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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$\begingroup$$\displaystyle \frac{8}{14}=\frac{4}{MP}\Rightarrow MP=7$
$\displaystyle \frac{14}{8}=\frac{7}{NP}\Rightarrow NP=4$
$\displaystyle\frac{14}{10}=\frac{7}{MN}\Rightarrow MN=5$
Perimeter$=MP+NP+MN=7+4+5=16$
$\endgroup$ 2 $\begingroup$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 $\frac{8+14+10}{2}=16$.
$\endgroup$ $\begingroup$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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