Chapter 6.5, Problem 42PPS

To determine

To prove:

The following statement with coordinate proof:

‘The segments joining the midpoints of the sides of rectangle form a rhombus.’

Explanation of Solution

Given information:

A rectangle

Proof:

Consider:

• $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is a rectangle
• Mid points of $▭\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ : $\mathrm{Q},\mathrm{R},\mathrm{S},\mathrm{T}$

  

Here, need to prove that the mid-point joining figure $\left(\mathrm{Q}\mathrm{R}\mathrm{S}\mathrm{T}\right)$ is rhombus.

Use midpoint formula to get the coordinates of the midpoint of a $▭\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ :

  $\begin{array}{l}\mathrm{Q}=\left(\frac{0+0}{2},\frac{\mathrm{b}+0}{2}\right)=\left(0,\frac{\mathrm{b}}{2}\right)\\ \mathrm{R}=\left(\frac{\mathrm{a}+0}{2},\frac{\mathrm{b}+\mathrm{b}}{2}\right)=\left(\frac{\mathrm{a}}{2},\mathrm{b}\right)\\ \mathrm{S}=\left(\frac{\mathrm{a}+\mathrm{a}}{2},\frac{\mathrm{b}+0}{2}\right)=\left(\mathrm{a},\frac{\mathrm{b}}{2}\right)\\ \mathrm{T}=\left(\frac{\mathrm{a}+0}{2},\frac{0+0}{2}\right)=\left(\frac{\mathrm{a}}{2},0\right)\end{array}$

Use the distance formula to calculate the lengths of figure $\left(\mathrm{Q}\mathrm{R}\mathrm{S}\mathrm{T}\right)$ :

  $\begin{array}{l}\mathrm{Q}\mathrm{R}=\sqrt{{\left(\frac{\mathrm{a}}{2}-0\right)}^2+{\left(\mathrm{b}-\frac{\mathrm{b}}{2}\right)}^2}=\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2}\\ \mathrm{R}\mathrm{S}=\sqrt{{\left(\mathrm{a}-\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}-\mathrm{b}\right)}^2}=\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2}\\ \mathrm{S}\mathrm{T}=\sqrt{{\left(\mathrm{a}-\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}-0\right)}^2}=\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2}\\ \mathrm{Q}\mathrm{T}=\sqrt{{\left(\frac{\mathrm{a}}{2}-0\right)}^2+{\left(0-\frac{\mathrm{b}}{2}\right)}^2}=\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2}\end{array}$

Get that:

  $\mathrm{Q}\mathrm{R}=\mathrm{R}\mathrm{S}=\mathrm{S}\mathrm{T}=\mathrm{Q}\mathrm{T}=\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2}$

That means:

  $\mathrm{Q}\mathrm{R}\cong \mathrm{R}\mathrm{S}\cong \mathrm{S}\mathrm{T}\cong \mathrm{Q}\mathrm{T}$

Therefore, the figure $\left(\mathrm{Q}\mathrm{R}\mathrm{S}\mathrm{T}\right)$ made by the midpoints of $▭\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is rhombus.

Hence proved