Chapter 10.1, Problem 35E

To determine

Explain the proof of the given statement.

Explanation of Solution

Given:

The given statement is that the midpoints of an isosceles trapezoid form a rhombus.

Calculation:

Draw a diagram for the given statement.

  

In the diagram, $PQRS$ is a trapezoid 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 $P$ , $Q,R$ and $S$ are $\left(-2a,0\right)$ , $\left(-a,2b\right)$ , $\left(a,2b\right)$ and $\left(2a,0\right)$ 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 $\left(PQ\right)$

  $A=\left(\frac{-2a-a}{2},\frac{0+2b}{2}\right)=\left(\frac{-3a}{2},b\right)$

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

  $B=\left(0,2b\right)$

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

  $C=\left(\frac{a+2a}{2},\frac{2b+0}{2}\right)=\left(\frac{3a}{2},b\right)$

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

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

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 the diagram

  $\begin{array}{l}AB=\sqrt{{\left[0-\left(-\frac{3a}{2}\right)\right]}^2+{\left(2b-b\right)}^2}=\sqrt{{\left(\frac{3a}{2}\right)}^2+b^2}\\ BC=\sqrt{{\left(\frac{3a}{2}-0\right)}^2+{\left(b-2b\right)}^2}=\sqrt{{\left(\frac{3a}{2}\right)}^2+b^2}\\ CD=\sqrt{-\frac{3a}{2}{\left[0-\left(\right)\right]}^2+{\left(0-b\right)}^2}=\sqrt{{\left(\frac{3a}{2}\right)}^2+b^2}\\ DA=\sqrt{{\left(-\frac{3a}{2}-0\right)}^2+{\left(b-0\right)}^2}=\sqrt{{\left(\frac{3a}{2}\right)}^2+b^2}\end{array}$

  $\begin{array}{l}AC=\sqrt{{\left[\frac{3a}{2}-\left(-\frac{3a}{2}\right)\right]}^2+{\left(b-b\right)}^2}=\sqrt{{\left(3a\right)}^2+0^2}=3a\\ BD=\sqrt{{\left(0-0\right)}^2+{\left(0-2b\right)}^2}=2b\end{array}$

From the above result

  $AB=BC=CD=DA$ and $AC\ne BD$ .

Hence $ABCD$ is a Rhombus.