Chapter 4.8, Problem 19PPS

To determine

To prove: The segments joining the midpoints of the legs of an isosceles triangle to the base vertices are congruent.

Explanation of Solution

Given:

Triangle is an isosceles triangle.

Calculation:

The figure can be drawn as:

  

The coordinates of R will be:

  $\begin{array}{l}=\left(\frac{0+\frac{1}{2}\mathrm{b}}{2},\frac{0+\mathrm{h}}{2}\right)\\ =\left(\frac{1}{4}\mathrm{b},\frac{\mathrm{h}}{2}\right)\end{array}$

The coordinates of S will be:

  $\begin{array}{l}=\left(\frac{\frac{1}{2}\mathrm{b}+\mathrm{b}}{2},\frac{0+\mathrm{h}}{2}\right)\\ =\left(\frac{3}{4}\mathrm{b},\frac{\mathrm{h}}{2}\right)\end{array}$

The length of AS will be:

  $\begin{array}{l}\mathrm{A}\mathrm{S}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ \mathrm{A}\mathrm{S}=\sqrt{{\left(\frac{3}{4}\mathrm{b}-0\right)}^2+{\left(\frac{\mathrm{h}}{2}-0\right)}^2}\\ \mathrm{A}\mathrm{S}=\sqrt{\frac{9}{16}{\mathrm{b}}^2+\frac{1}{4}{\mathrm{h}}^2}\\ \text{Hence,}\\ \mathrm{A}\mathrm{S}=\mathrm{B}\mathrm{R}\\ \text{implies that,}\\ \overline{\mathrm{A}\mathrm{S}}\cong \overline{\mathrm{B}\mathrm{R}}\end{array}$