Chapter 4.8, Problem 22PPS

To determine

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

Explanation of Solution

The figure can be drawn as:

  

The coordinates of S will be:

  $\begin{array}{l}=\left(\frac{0+\mathrm{a}}{2},\frac{0+\mathrm{h}}{2}\right)\\ =\left(\frac{\mathrm{a}}{2},\frac{\mathrm{h}}{2}\right)\end{array}$

The coordinates of T will be:

  $\begin{array}{l}=\left(\frac{\mathrm{a}+\mathrm{b}}{2},\frac{0+\mathrm{h}}{2}\right)\\ =\left(\frac{\mathrm{a}+\mathrm{b}}{2},\frac{\mathrm{h}}{2}\right)\end{array}$

The length of BC will be:

  $\begin{array}{l}\mathrm{B}\mathrm{C}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ \mathrm{B}\mathrm{C}=\sqrt{{\left(\mathrm{b}-0\right)}^2+{\left(0-\mathrm{h}\right)}^2}\\ \mathrm{B}\mathrm{C}=\sqrt{{\mathrm{b}}^2+{\mathrm{h}}^2}\end{array}$

The length of AB will be:

  $\begin{array}{l}\mathrm{A}\mathrm{B}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ \mathrm{A}\mathrm{B}=\sqrt{{\left(\mathrm{b}-0\right)}^2+{\left(0-0\right)}^2}\\ \mathrm{A}\mathrm{B}=\sqrt{{\mathrm{b}}^2+0}\\ \mathrm{A}\mathrm{B}=\mathrm{b}\end{array}$

The length of ST will be:

  $\begin{array}{l}\mathrm{S}\mathrm{T}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ \mathrm{S}\mathrm{T}=\sqrt{{\left(\frac{\mathrm{a}+\mathrm{b}}{2}-\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{h}}{2}-\frac{\mathrm{h}}{2}\right)}^2}\\ \mathrm{S}\mathrm{T}=\sqrt{\frac{1}{4}{\mathrm{b}}^2+0}\\ \mathrm{S}\mathrm{T}=\frac{1}{2}\mathrm{b}\\ \text{Hence,}\\ \mathrm{S}\mathrm{T}=\frac{1}{2}\left(\mathrm{A}\mathrm{B}\right)\end{array}$