Chapter 6.6, Problem 33PPS

To determine

To prove:

The following statement with coordinate proof:

Theorem 6.24: $\overline{\mathrm{E}\mathrm{F}}\parallel \overline{\mathrm{A}\mathrm{B}}$ and $\overline{\mathrm{E}\mathrm{F}}\parallel \overline{\mathrm{D}\mathrm{C}}$ and $\overline{\mathrm{E}\mathrm{F}}=\frac{1}{2}\left(\mathrm{A}\mathrm{B}+\mathrm{D}\mathrm{C}\right)$

Explanation of Solution

Given information:

  $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is a trapezoid with median $\overline{\mathrm{E}\mathrm{F}}$

Proof:

According to the definition of trapezoid $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ , $\mathrm{E}$ and $\mathrm{F}$ is midpoint of $\overline{\mathrm{A}\mathrm{D}}$ and $\overline{\mathrm{B}\mathrm{C}}$

  

Use midpoint formula to get the coordinates of the midpoint of a $\mathrm{E}$ and $\mathrm{F}$ :

  $\begin{array}{l}\mathrm{E}=\left(\frac{\mathrm{a}+0}{2},\frac{\mathrm{d}+0}{2}\right)=\left(\frac{\mathrm{a}}{2},\frac{\mathrm{d}}{2}\right)\\ \mathrm{F}=\left(\frac{\mathrm{a}+\mathrm{b}+\mathrm{a}+\mathrm{b}+\mathrm{c}}{2},\frac{\mathrm{d}+0}{2}\right)=\left(\frac{2\mathrm{a}+2\mathrm{b}+\mathrm{c}}{2},\frac{\mathrm{d}}{2}\right)\end{array}$

So, slope of $\overline{\mathrm{A}\mathrm{B}}=0$ and slope of $\overline{\mathrm{E}\mathrm{F}}=0$ and slope of $\overline{\mathrm{D}\mathrm{C}}=0$ . Thus, $\overline{\mathrm{E}\mathrm{F}}\parallel \overline{\mathrm{A}\mathrm{B}}$ and $\overline{\mathrm{E}\mathrm{F}}\parallel \overline{\mathrm{D}\mathrm{C}}$ .

Use the distance formula to calculate the lengths:

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

Now, take the equation: $\overline{\mathrm{E}\mathrm{F}}=\frac{1}{2}\left(\mathrm{A}\mathrm{B}+\mathrm{D}\mathrm{C}\right)$

Use its right hand side as:

  $\begin{array}{l}\frac{1}{2}\left(\mathrm{A}\mathrm{B}+\mathrm{D}\mathrm{C}\right)=\frac{1}{2}\left[\mathrm{b}+\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\right]\\ \frac{1}{2}\left(\mathrm{A}\mathrm{B}+\mathrm{D}\mathrm{C}\right)=\frac{1}{2}\left(\mathrm{a}+2\mathrm{b}+\mathrm{c}\right)\\ \frac{1}{2}\left(\mathrm{A}\mathrm{B}+\mathrm{D}\mathrm{C}\right)=\frac{1}{2}\left[\mathrm{b}+\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\right]\\ \frac{1}{2}\left(\mathrm{A}\mathrm{B}+\mathrm{D}\mathrm{C}\right)=\frac{\mathrm{a}+2\mathrm{b}+\mathrm{c}}{2}\\ \frac{1}{2}\left(\mathrm{A}\mathrm{B}+\mathrm{D}\mathrm{C}\right)=\overline{\mathrm{E}\mathrm{F}}\end{array}$

Hence proved