Chapter 7.4, Problem 52HP

To determine

To compare: The triangle proportionality theorem and the triangle midsegment theorem.

Explanation of Solution

Given information:

The triangle proportionality theorem and the triangle midsegment theorem.

The triangle proportionality theorem is, “If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional length”.

From the above statement, $\overline{BD}\left|\right|\overline{AE}$ .

Consider the following figure,

  

From AA similarity postulate, $\Delta ACE\sim \Delta BCD$ and,

  $\begin{array}{l}\frac{CA}{CB}=\frac{CE}{CD}\\ \frac{CB+BA}{CB}=\frac{CD+DE}{CD}\\ 1+\frac{BA}{CB}=1+\frac{DE}{CD}\\ \frac{BA}{CB}=\frac{DE}{CD}\end{array}$

Hence, if a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional length.

The triangle midsegment 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.