#### Answer

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

#### 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: $~RS = x = \frac{b^2}{b+c} = \frac{bc}{b+c}$