#### 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, $$\overline{RA}\cong\overline{RC}$$