Chapter 6.4, Problem 42PPS

To determine

To prove:

The following statement with the coordinate proof:

‘The diagonals of a rectangle are congruent.’

Explanation of Solution

Given information:

Rectangle

Proof:

Let’s draw a rectangle $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ in the coordinate plane. Name point $\mathrm{D}$ at the origin. Consider the length and height of as rectangle $\mathrm{a}$ and $\mathrm{b}$ 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 rectangle. Here need to prove $\overline{\mathrm{A}\mathrm{C}}\cong \overline{\mathrm{D}\mathrm{B}}$ where $\overline{\mathrm{A}\mathrm{C}}$ and $\overline{\mathrm{D}\mathrm{B}}$ are diagonal of the rectangle.

  

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

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

Here, get that:

The length of diagonals $\overline{\mathrm{A}\mathrm{C}}$ and $\overline{\mathrm{D}\mathrm{B}}$ is same $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$ which implies that diagonals of the rectangle are congruent as $\overline{\mathrm{A}\mathrm{C}}\cong \overline{\mathrm{D}\mathrm{B}}$ .

Hence proved