Line segment joining mid points of any two sides of a triangle is parallel to the third side and has a length half of the length of the third side.

**Proof:**

$PM$ = $MR$ (given)

$PN$ = $NQ$ (given)

So, $PR$ = $PM + MR$ = $2\ PM$

$PQ$ = $PN + NQ$ = $2\ PN$

In triangle $PMN$ and triangle $PRQ$, angle $P$ is common

$\frac{PM}{PR}$ = $\frac{PN}{PQ}$ = $\frac{1}{2}$

As because the angle is same and the sides are in proportion we can say triangle $PMN$ and triangle $PRQ$ are similar triangles by side angle side postulate.

As $MN$ and $RQ$ are parts of the similar triangle $PMN$ and $PRQ$

Hence, $\frac{MN}{RQ}$ = $\frac{PM}{PR}$ = $\frac{PN}{PQ}$ = $\frac{1}{2}$ or $MN$ = $\frac{1}{2}$ $RQ$

Also, angle $PMN$ and angle $PRQ$ are corresponding angles of the similar triangle $PMN$ and $PRQ$ and we know that corresponding angles are always equal. So, angle $PMN$ = angle $PRQ$

The line segments $MN$ and $RQ$ are cut by the transversal $PR$ and the angle $PMN$ is equal to angle $PRQ$. Therefore, according to the parallel and transversal property we derive $MN$ is parallel to $RQ$. Hence, proved.