Chapter 13.9, Problem 7WE

To determine

To prove: that the figure formed by joining, in order, the midpoints of the sides of quadrilateral $ROST$ is a parallelogram.

Explanation of Solution

Given information:

  $ROST$ is a quadrilateral with given vertices $R(2b,2c),T(2d,2e),S(2a,0),O(0,0)$ .

  

Calculation:

Consider the mid points $M_1,M_2,M_3,M_4$ of sides $OS,ST,TR,RO$ respectively and joining so that,

  

  $\begin{array}{l}{M}_1=\left(\frac{2a+0}{2},\frac{0+0}{2}\right)=(a,0)\\ M_2=\left(\frac{2(a+d)}{2},\frac{2e}{2}\right)=\left(a+d,e\right)\\ M_3=\left(\frac{2b+2d}{2},\frac{2c+2e}{2}\right)=(b+d,c+e)\\ M_4=\left(\frac{2b+0}{2},\frac{2c+0}{2}\right)=(b,c)\end{array}$

We have to show, $\overline{M_1{M}_2}=\overline{M_3{M}_4}$ , and $\overline{M_1{M}_4}=\overline{M_3{M}_2}$ .

Now,

  $\begin{array}{l}\overline{M_1{M}_2}=\sqrt{{(a+d-a)}^2+{(e-0)}^2}=\sqrt{d^2+e^2}\\ \overline{M_3{M}_4}=\sqrt{{(b-b-d)}^2+{(c-c-e)}^2}=\sqrt{d^2+e^2}\end{array}$

That is $\overline{M_1{M}_2}=\overline{M_3{M}_4}$ . Again,

  $\begin{array}{l}\overline{M_1{M}_4}=\sqrt{c^2+{(b-a)}^2}\\ \overline{M_1{M}_2}=\sqrt{{(c+e-e)}^2+{(b+d-a-d)}^2}=\sqrt{c^2+{(b-a)}^2}\end{array}$

Implies $\overline{M_1{M}_4}=\overline{M_3{M}_2}$ .

Hence $M_1{M}_2{M}_3{M}_4$ represents a parallelogram is proved.