Chapter 7.4, Problem 32PPS

To determine

To write: The two-column proof of Theorem $7.7$ .

Explanation of Solution

Given information:

The given Theorem is, “A mid-segment of a triangle is parallel to one side of the triangle and its length is one half the length of that side”.

The given Theorem is, “A mid-segment of a triangle is parallel to one side of the triangle and its length is one half the length of that side”.

Consider the following figure,

  

From the above statement, $\overline{DE}$ is a mid-segment of $\Delta ABC$ . From mid-point theorem,

  $\begin{array}{l}AD=DB\\ AE=EC\end{array}$

From the figure,

  $\begin{array}{l}AB=AD+DB\\ AB=AD+AD\\ AB=2AD\\ \frac{AB}{AD}=2\end{array}$

And,

  $\begin{array}{l}AC=AE+EC\\ AC=AE+AE\\ AC=2AE\\ \frac{AC}{AE}=2\end{array}$

So, $\frac{AB}{AD}=\frac{AC}{AE}$ .

From SAS similarity, $\Delta ADE\sim \Delta ABC$ and $\angle ADE=\angle ABC$ , so $\overline{DE}\left|\right|\overline{BC}$ .

From $\Delta ADE\sim \Delta ABC$

  $\begin{array}{l}\frac{BC}{DE}=\frac{AB}{AD}\\ \frac{BC}{DE}=2\\ DE=\frac{1}{2}BC\end{array}$

Hence, a mid-segment of a triangle is parallel to one side of the triangle and its length is one half the length of that side.