Chapter 1.1, Problem 56E

To determine

To show:If the midpoints of consecutive sides of any quadrilateral are connected, then it formed a parallelogram.

Explanation of Solution

Given information:The shape is a quadrilateral and midpoints of their consecutive sides.

Calculation:

Assume that the vertices of any quadrilateral are $A\left(x_1,y_1\right)$ , $B\left(x_2,y_2\right)$ , $C\left(x_3,y_3\right)$ and $D\left(x_4,y_4\right)$ .

The midpoint of line $\overline{AB}$ is:

  $M_1=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

The midpoint of line $\overline{BC}$ is:

  $M_2=\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$

The midpoint of line $\overline{CD}$ is:

  $M_3=\left(\frac{x_3+x_4}{2},\frac{y_3+y_4}{2}\right)$

The midpoint of line $\overline{DA}$ is:

  $M_4=\left(\frac{x_4+x_1}{2},\frac{y_4+y_1}{2}\right)$

To show that the figure drawn by joining the midpoints of the consecutive sides of any quadrilateral is a parallelogram, show that the slopes of opposite sides are equal.

The slope between the consecutive midpoints $M_1$ and $M_2$ is:

  $\begin{array}{c}{m}_{M_1{M}_2}=\frac{\frac{y_2+y_3}{2}-\frac{y_1+y_2}{2}}{\frac{x_2+x_3}{2}-\frac{x_1+x_2}{2}}\\ =\frac{y_3-y_1}{x_3-x_1}\end{array}$

The slope between the consecutive midpoints $M_3$ and $M_4$ is:

  $\begin{array}{c}{m}_{M_3{M}_4}=\frac{\frac{y_4+y_1}{2}-\frac{y_3+y_4}{2}}{\frac{x_4+x_1}{2}-\frac{x_3+x_4}{2}}\\ =\frac{y_1-y_3}{x_1-x_3}\\ =\frac{y_3-y_1}{x_3-x_1}\end{array}$

It can be said that the opposite two sides have the same slope. It is known that if two sides have same slope, they are parallel. The line that join the midpoints $M_1$ and $M_2$ is parallel to the line that join the midpoints $M_3$ and $M_4$ .

Similarly, $m_{\overline{M_2{M}_3}}=m_{\overline{M_4{M}_1}}$ .

Hence, it is proved that the midpoints of consecutive sides of any quadrilateral are connected then it formed a parallelogram.