Chapter 6.6, Problem 64PPS

To determine

To prove:

The following statement with the coordinate proof:

  $\overline{\mathrm{X}\mathrm{Y}}\parallel \overline{\mathrm{A}\mathrm{B}}$ and $\overline{\mathrm{X}\mathrm{Y}}\parallel \overline{\mathrm{D}\mathrm{C}}$

Explanation of Solution

Given information:

Isosceles trapezoid $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ with median $\overline{\mathrm{X}\mathrm{Y}}$

Proof:

Consider,

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

Length of longer base: $\mathrm{a}+2\mathrm{b}\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$

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: $\mathrm{A}\left(-\mathrm{b},\mathrm{c}\right),\mathrm{B}\left(\mathrm{a}+\mathrm{b},\mathrm{c}\right),\mathrm{C}\left(\mathrm{a},0\right)$

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

is drawn as below:

  

Here, the midpoint of $\overline{\mathrm{A}\mathrm{D}}$ is $\mathrm{X}$ , so its coordinates are $\left(\frac{-\mathrm{b}}{2},\frac{\mathrm{c}}{2}\right)$ and the midpoint of $\overline{\mathrm{B}\mathrm{C}}$ is $\mathrm{Y}$ , so its coordinates are $\left(\frac{2\mathrm{a}+\mathrm{b}}{2},\frac{\mathrm{c}}{2}\right)$ . Use the slope formula to get the slope values of bases and median:

  $\begin{array}{l}{\mathrm{m}}_{\mathrm{A}\mathrm{B}}=\frac{\mathrm{c}-\mathrm{c}}{\mathrm{a}+\mathrm{b}-\left(-\mathrm{b}\right)}=0\\ {\mathrm{m}}_{\mathrm{D}\mathrm{C}}=\frac{0-0}{\mathrm{a}-0}=0\\ {\mathrm{m}}_{\mathrm{X}\mathrm{Y}}=\frac{\left(\frac{\mathrm{c}}{2}-\frac{\mathrm{c}}{2}\right)}{\left[\frac{2\mathrm{a}+\mathrm{b}}{2}-\left(\frac{-\mathrm{b}}{2}\right)\right]}=0\end{array}$

So, get that:

  $\overline{\mathrm{X}\mathrm{Y}}\parallel \overline{\mathrm{A}\mathrm{B}}$ and $\overline{\mathrm{X}\mathrm{Y}}\parallel \overline{\mathrm{D}\mathrm{C}}$

Hence proved