Chapter 6.6, Problem 63PPS

To determine

To prove:

The following statement with the coordinate proof:

  $\overline{\mathrm{B}\mathrm{D}}\cong \overline{\mathrm{A}\mathrm{C}}$

Explanation of Solution

Given information:

Isosceles trapezoid $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ with $\overline{\mathrm{A}\mathrm{D}}\cong \overline{\mathrm{B}\mathrm{C}}$

Proof:

Consider,

Length of shorter base: $\mathrm{b}\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

Length of longer base: $\mathrm{a}\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

Height: $\mathrm{c}\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

Therefore, the vertices of $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ are: $\mathrm{B}\left(\mathrm{a},0\right),\mathrm{C}\left(\mathrm{a}-\mathrm{b},\mathrm{c}\right),\mathrm{D}\left(\mathrm{b},\mathrm{c}\right)$

Isosceles trapezoid $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$

is drawn as below:

  

Use the distance formula to get the value of diagonals of isosceles trapezoid $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ :

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

So, get that:

  $\mathrm{B}\mathrm{D}=\mathrm{A}\mathrm{C}$

Thus, $\overline{\mathrm{B}\mathrm{D}}\cong \overline{\mathrm{A}\mathrm{C}}$

Hence proved