Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 5 - Section 5.3 - Proving Lines Parallel - Exercises - Page 233: 43


$RS = \frac{bc}{b+c}$

Work Step by Step

$\angle BAC \cong \angle BRS$ because opposite sides of a rhombus are parallel. $\angle ABC \cong \angle RBS$ because they are the same angle. Therefore, $\triangle ABC$ ~ $\triangle RBS$ Also note that the length of $RS = AR = x$ since $ARST$ is a rhombus. That is, all sides of a rhombus have the same length. Then: $\frac{RB}{RS} = \frac{AB}{AC}$ $\frac{c-x}{x} = \frac{b}{c}$ $c^2-cx = bx$ $bx+cx = c^2$ $x = \frac{c^2}{b+c}$ By the same argument, we can show that $\triangle ABC$ ~ $\triangle TCS$, which would lead to $x = \frac{b^2}{b+c}$ Therefore, $b^2 = c^2$ and so $b = c$ Then: $~RS = x = \frac{b^2}{b+c} = \frac{bc}{b+c}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.