Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 3 - Section 3.1 - Congruent Triangles - Exercises: 41

Answer

- Prove that $\triangle RSA\cong\triangle RSB$ to show that $\overline{RA}\cong\overline{RB}$. - Prove that $\triangle RTB\cong\triangle RTC$ to show that $\overline{RB}\cong\overline{RC}$. - Now, it follows that $\overline{RA}\cong\overline{RC}$

Work Step by Step

*STRATEGY: - Prove that $\triangle RSA\cong\triangle RSB$ to show that $\overline{RA}\cong\overline{RB}$. - Prove that $\triangle RTB\cong\triangle RTC$ to show that $\overline{RB}\cong\overline{RC}$. - Now, it follows that $\overline{RA}\cong\overline{RC}$ 1) Prove that $\triangle RSA\cong\triangle RSB$ Since $\overline{RS}$ is the perpendicular bisector of $\overline{AB}$, it follows that $\overline{SA}\cong\overline{SB}$ and $\angle RSA\cong\angle RSB$ (since both angles are $90^{\circ}$) Furthermore, by Identity, we see that $\overline{RS}\cong\overline{RS}$. Therefore, by method SAS, $\triangle RSA\cong\triangle RSB$. So, $\overline{RA}\cong\overline{RB}$. 2) Prove that $\triangle RTB\cong\triangle RTC$ Similarly, since $\overline{RT}$ is the perpendicular bisector of $\overline{BC}$, it follows that $\overline{TB}\cong\overline{TC}$ and $\angle RTB\cong\angle RTC$ (since both angles are $90^{\circ}$) Furthermore, by Identity, we see that $\overline{RT}\cong\overline{RT}$. Therefore, by method SAS, $\triangle RTB\cong\triangle RTC$. So, $\overline{RB}\cong\overline{RC}$. 3) From the results from 1) and 2), it follows that $$\overline{RA}\cong\overline{RB}\cong\overline{RC}$$ So, $$\overline{RA}\cong\overline{RC}$$
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