Chapter 10.1, Problem 28E

To determine

Prove that the midpoint theorem analytically.

Explanation of Solution

Given:

The line segment joining the middle points of two sides of a triangle is equal to half of the third side.

Calculation:

Draw a triangle for the given statement.

  

In triangle, points $D$ and $E$ are the midpoints of the segment $AB$ and $AC$ respectively.

Let the coordinates of the points $A,B$ and $C$ are $\left(x_1,y_1\right)$ , $\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ respectively.

If the coordinates of $p_1$ and $p_2$ are $\left(p_1,q_1\right)$ and $\left(p_2,q_2\right)$ , respectively, then the midpoint of $\overline{p_1{p}_2}$ has coordinates $\left(\frac{p_1+p_2}{2},\frac{q_1+q_2}{2}\right)$ .

Coordinates of $D$

  $=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Coordinates of $E$

  $=\left(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\right)$

The distance, $d$ units, between two points with coordinates $\left(p_1,q_1\right)$ and $\left(p_2,q_2\right)$ is given by $d=\sqrt{{\left(p_2-p_1\right)}^2+{\left(q_2-q_1\right)}^2}$ .

For the given points $B$ and $C$ .

  $\begin{array}{l}{p}_1=x_2,p_2=x_3\\ q_1=y_2,q_2=y_3\end{array}$

  $BC=\sqrt{{\left(x_3-x_2\right)}^2+{\left(y_3-y_2\right)}^2}$

For the given points $D$ and $E$ .

  $\begin{array}{l}{p}_1=\frac{x_1+x_2}{2},p_2=\frac{x_1+x_3}{2}\\ q_1=\frac{y_1+y_2}{2},q_2=\frac{y_1+y_3}{2}\end{array}$

  $\begin{array}{l}DE=\sqrt{{\left(\frac{x_1+x_3}{2}-\frac{x_1+x_2}{2}\right)}^2+{\left(\frac{y_1+y_3}{2}-\frac{y_1+y_2}{2}\right)}^2}\\ =\sqrt{{\left(\frac{x_3-x_2}{2}\right)}^2+{\left(\frac{y_3-y_2}{2}\right)}^2}\\ =\frac{1}{2}\sqrt{{\left(x_3-x_2\right)}^2+{\left(y_3-y_2\right)}^2}\end{array}$

Substitute the value of $\sqrt{{\left(x_3-x_2\right)}^2+{\left(y_3-y_2\right)}^2}$ by $BC$ .

  $\Rightarrow DE=\frac{1}{2}BC$

Hence proved