Prove that,if the midpoints of the sides of a quadrilateral are joined in order,we get a parallelogram.

(N/A) Let the vertices of the quadrilateral $\square ABCD$ be $A(x_1, y_1)$,$B(x_2, y_2)$,$C(x_3, y_3)$,and $D(x_4, y_4)$.
Let $P, Q, R$,and $S$ be the midpoints of sides $\overline{AB}$,$\overline{BC}$,$\overline{CD}$,and $\overline{DA}$ respectively.
Using the midpoint formula,the coordinates are:
$P = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$,$Q = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)$,$R = \left(\frac{x_3+x_4}{2}, \frac{y_3+y_4}{2}\right)$,$S = \left(\frac{x_4+x_1}{2}, \frac{y_4+y_1}{2}\right)$.
$A$ quadrilateral is a parallelogram if its diagonals bisect each other,meaning they share the same midpoint.
Midpoint of diagonal $\overline{PR}$:
$= \left( \frac{\frac{x_1+x_2}{2} + \frac{x_3+x_4}{2}}{2}, \frac{\frac{y_1+y_2}{2} + \frac{y_3+y_4}{2}}{2} \right) = \left( \frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4} \right)$.
Midpoint of diagonal $\overline{QS}$:
$= \left( \frac{\frac{x_2+x_3}{2} + \frac{x_4+x_1}{2}}{2}, \frac{\frac{y_2+y_3}{2} + \frac{y_4+y_1}{2}}{2} \right) = \left( \frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4} \right)$.
Since the midpoints of diagonals $\overline{PR}$ and $\overline{QS}$ are identical,the diagonals bisect each other.
Therefore,$\square PQRS$ is a parallelogram.