Chapter P.4, Problem 71E

Solution

To show that in the given triangle, the line segment joining the midpoints of two sides is parallel to the third side and is half as long.

Given:

The figure,

  

Concept Used:

• The slope-intercept form of a line is $y=mx+b$, where m is the slope of the line and b is the y -intercept of the line.
• Slope of the line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$
• The coordinates of the line segment joining the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Calculation:

In order to show that in any triangle, the line segment joining the midpoints of two sides is parallel to the third side, note that the coordinates of the points A are $\left(\frac{b}{2},\frac{c}{2}\right)$ and the point B has coordinates $\left(\frac{a+b}{2},\frac{c}{2}\right)$ .

Since the y -coordinate of the points A and B is same, it implies that the line segment AB is a parallel line and hence parallel to the third side.

The distance between points A and B is:

  $\begin{array}{c}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(\frac{a+b}{2}-\frac{b}{2}\right)}^2+{\left(\frac{c}{2}-\frac{c}{2}\right)}^2}\\ =\sqrt{{\left(\frac{a+b-b}{2}\right)}^2+{\left(0\right)}^2}\\ =\sqrt{{\left(\frac{a}{2}\right)}^2+0}\\ =\frac{a}{2}\end{array}$

Since the length of the third side is a , it implies that the length of the line segment AB is half as long the third side.