Chapter 6.4, Problem 43PPS

To determine

To prove:

The following statement with the coordinate proof:

‘The diagonals of a parallelogram are congruent, and then it is a rectangle.’

Explanation of Solution

Given information:

The diagonals of a parallelogram are congruent Proof:

Let’s draw a parallelogram $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ in the coordinate plane. Name point $\mathrm{D}$ at the origin. Consider the length and height of as rectangle $\mathrm{a}$ and $\mathrm{c}$ respectively. The remaining vertices of rectangle are named as $\mathrm{A}\left(0,\mathrm{b}\right),\mathrm{B}\left(\mathrm{a},\mathrm{b}\right),\mathrm{C}\left(\mathrm{a},0\right)$ .

It is given that $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is parallelogram which has congruent diagonals as $\overline{\mathrm{A}\mathrm{C}}\cong \overline{\mathrm{D}\mathrm{B}}$ . Here need to prove $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is rectangle.

  

Use the distance formula to find the lengths of the diagonals as:

  $\begin{array}{l}\mathrm{A}\mathrm{C}=\sqrt{{\left(\mathrm{a}+\mathrm{b}-0\right)}^2+{\left(\mathrm{c}-0\right)}^2}=\sqrt{{\left(\mathrm{a}+\mathrm{b}\right)}^2+{\mathrm{c}}^2}\\ \mathrm{B}\mathrm{D}=\sqrt{{\left(\mathrm{b}-\mathrm{a}\right)}^2+{\left(\mathrm{c}-0\right)}^2}=\sqrt{{\left(\mathrm{b}-\mathrm{a}\right)}^2+{\mathrm{c}}^2}\end{array}$

Here, know that:

  $\begin{array}{l}\mathrm{A}\mathrm{C}=\mathrm{B}\mathrm{D}\\ \sqrt{{\left(\mathrm{a}+\mathrm{b}\right)}^2+{\mathrm{c}}^2}=\sqrt{{\left(\mathrm{b}-\mathrm{a}\right)}^2+{\mathrm{c}}^2}\\ {\left(\mathrm{a}+\mathrm{b}\right)}^2+{\mathrm{c}}^2={\left(\mathrm{b}-\mathrm{a}\right)}^2+{\mathrm{c}}^2\\ {\left(\mathrm{a}+\mathrm{b}\right)}^2={\left(\mathrm{b}-\mathrm{a}\right)}^2\\ {\mathrm{a}}^2+2\mathrm{a}\mathrm{b}+{\mathrm{b}}^2+{\mathrm{c}}^2={\mathrm{b}}^2-2\mathrm{a}\mathrm{b}+{\mathrm{a}}^2+{\mathrm{c}}^2\\ 2\mathrm{a}\mathrm{b}=-2\mathrm{a}\mathrm{b}\\ 4\mathrm{a}\mathrm{b}=0\\ \mathrm{a}=0\mathrm{O}\mathrm{R}\text{}\mathrm{b}=0\end{array}$

Because $\mathrm{A}$ and

  $\mathrm{B}$ different points $\mathrm{a}\ne 0$ so, $\mathrm{b}=0$ . It means that the slope of $\overline{\mathrm{A}\mathrm{D}}$ and $\overline{\mathrm{A}\mathrm{B}}$ is undefined and zero so, $\overline{\mathrm{A}\mathrm{D}}\perp \overline{\mathrm{A}\mathrm{B}}$ . So, $\angle \mathrm{D}\mathrm{A}\mathrm{B}$ is right angle and $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is rectangle.

Hence proved