Chapter 10.1, Problem 32E

To determine

Explain the proof of the given statement.

Explanation of Solution

Given:

The given statement is that the midpoints of a quadrilaterals form a parallelogram.

Calculation:

Draw a diagram for the given statement.

  

In the diagram, $PQRS$ is the quadrilateral and $A,B,C$ , $D$ are the midpoints of the segment $\overline{PQ},\overline{QR},\overline{RS},\overline{SP}$ respectively.

Let the coordinates of the points $P$ (origin), $Q,R$ and $S$ are $\left(0,0\right)$ , $\left(0,b\right)$ , $\left(c,d\right)$ and $\left(a,0\right)$ respectively.

If the coordinates of $p_1$ and $p_2$ are $\left(p_1,q_1\right)$ and $\left(p_2,q_2\right)$ , respectively, then the midpoint of $\overline{p_1{p}_2}$ has coordinates $\left(\frac{p_1+p_2}{2},\frac{q_1+q_2}{2}\right)$ .

Coordinates of the midpoint $\left(PQ\right)$

  $=\left(\frac{0+0}{2},\frac{0+b}{2}\right)=\left(0,\frac{b}{2}\right)$

Coordinates of the midpoint $\left(QR\right)$

  $=\left(\frac{0+c}{2},\frac{b+d}{2}\right)=\left(\frac{c}{2},\frac{b+d}{2}\right)$

Coordinates of the midpoint $\left(RS\right)$

  $=\left(\frac{c+a}{2},\frac{d+0}{2}\right)=\left(\frac{c+a}{2},\frac{d}{2}\right)$

Coordinates of the midpoint $\left(QR\right)$

  $=\left(\frac{a+0}{2},\frac{0+0}{2}\right)=\left(\frac{a}{2},0\right)$

The distance, $d$ units, between two points with coordinates $\left(p_1,q_1\right)$ and $\left(p_2,q_2\right)$ is given by $d=\sqrt{{\left(p_2-p_1\right)}^2+{\left(q_2-q_1\right)}^2}$ .

In triangles $\Delta PQR$ and $\Delta AQB$ .

  $\begin{array}{l}PQ=\sqrt{{\left(0-0\right)}^2+{\left(b-0\right)}^2}=b\\ AQ=\sqrt{{\left(0-0\right)}^2+{\left(\frac{b}{2}-0\right)}^2}=\frac{b}{2}\\ QR=\sqrt{{\left(0-c\right)}^2+{\left(b-d\right)}^2}=\sqrt{c^2+{\left(b-d\right)}^2}\\ BQ=\sqrt{{\left(0-\frac{c}{2}\right)}^2+{\left(b-\frac{b+d}{2}\right)}^2}=\sqrt{{\frac{c}{4}}^2+{\left(\frac{b-d}{4}\right)}^2}=\frac{1}{2}\sqrt{c^2+{\left(b-d\right)}^2}\end{array}$

  $\begin{array}{l}AQ=\frac{1}{2}PQ\\ BQ=\frac{1}{2}QR\end{array}$

Now the $\Delta PQR\sim \Delta AQB$

Then,

  $AB=\frac{1}{2}PR$

From midpoint theorem

  $AB\parallel PR---\left(i\right)$

In triangles $\Delta PSR$ and $\Delta CSD$ .

  $\begin{array}{l}SR=\sqrt{{\left(a-c\right)}^2+{\left(0-d\right)}^2}=\sqrt{{\left(a-c\right)}^2+d^2}\\ SC=\sqrt{{\left(a-\frac{c+a}{2}\right)}^2+{\left(0-\frac{d}{2}\right)}^2}=\sqrt{{\left(\frac{a-c}{4}\right)}^2+{\left(\frac{d}{2}\right)}^2}=\frac{1}{2}\sqrt{{\left(a-c\right)}^2+d^2}\\ SP=\sqrt{{\left(a-0\right)}^2+{\left(0-0\right)}^2}=a\\ SD=\sqrt{{\left(0-\frac{a}{2}\right)}^2+{\left(0-0\right)}^2}=\frac{a}{2}\end{array}$

  $\begin{array}{l}SC=\frac{1}{2}RS\\ SD=\frac{1}{2}PS\end{array}$

Now the $\Delta PSR\sim \Delta CSD$

Then,

  $CD=\frac{1}{2}PR$

From midpoint theorem

  $CD\parallel PR---\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$ .

  $AB\parallel CD$

Similarly $AD\parallel BC$ .

Hence $▱ABCD$ is a parallelogram.