Chapter 9.1, Problem 11P

To determine

To prove: The line joining the midpoints of two sides of a triangle is half to the length of its third side.

Explanation of Solution

Given information: The points are $P_1\left(a,b\right)\text{and}{P}_2\left(c,d\right)$ .

Formula used: $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ and $M{P}_1=M{P}_2=\frac{1}{2P_1{P}_2}$

Proof: Get the midpoints first.

  $M=\left(\frac{a+c}{2},\frac{b+d}{2}\right)$

Now, calculate $M{P}_1$ ,

  $\begin{array}{l}M{P}_1=\sqrt{{\left(\frac{\alpha +c}{2}-a\right)}^2+{\left(\frac{b+d}{2}-b\right)}^2}\\ =\sqrt{{\left(\frac{c}{2}-\frac{a}{2}\right)}^2+{\left(\frac{d}{2}-\frac{b}{2}\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}+\frac{c^2}{4}+\frac{d^2}{4}-\frac{2ac}{4}-\frac{2bd}{4}}\\ =\left(\frac{\sqrt{a^2+b^2+c^2+d^2-2ac-2bd}}{2}\right)\end{array}$

Similarly, $M{P}_2$ ,

  $\begin{array}{l}M{P}_2=\sqrt{{\left(\frac{\alpha +c}{2}-c\right)}^2+{\left(\frac{b+d}{2}-d\right)}^2}\\ =\sqrt{{\left(\frac{a}{2}-\frac{c}{2}\right)}^2+{\left(\frac{b}{2}-\frac{d}{2}\right)}^2}\\ =\sqrt{\frac{a^2}{4}+\frac{b^2}{4}+\frac{c^2}{4}+\frac{d^2}{4}-\frac{2ac}{4}-\frac{2bd}{4}}\\ =\left(\frac{\sqrt{a^2+b^2+c^2+d^2-2ac-2bd}}{2}\right)\end{array}$

Now,

  $\begin{array}{l}\therefore \frac{1}{2P_1{P}_2}=\left(\frac{\sqrt{{\left(a-c\right)}^2+{\left(b-d\right)}^2}}{2}\right)\\ =\left(\frac{\sqrt{a^2-2ac+c^2+b^2-2bc+d^2}}{2}\right)\\ =\left(\frac{\sqrt{a^2+b^2+c^2+d^2-2ac-2bd}}{2}\right)\end{array}$

Hence, $M{P}_1=M{P}_2=\frac{1}{2P_1{P}_2}$