Let $r$ be the inradius of triangle $ABC$. Then the area of triangle $ABC$ is $\frac{r(AB + AC + BC)}{2} = 2r\cdot BC$. Thus the length of the altitude from $A$ to $BC$ is $\frac{2(2r \cdot BC)}{BC} = 4r$. Since $KLCB$ is a trapezoid where $KL$ and $BC$ are parallel line segments both tangent to a circle with radius $r$, the distance between segments $KL$ and $BC$ is $2r$. So the length of the altitude from $A$ to $KL$ is $4r - 2r = 2r$. Since $\triangle AKL \sim \triangle ABC$, $\frac{KL}{BC} = \frac{1}{2}$, which implies that $K$ and $L$ are the midpoints of $AB$ and $AC$ respectively. Hence, $AT = TL$. By Menelaus' theorem, $\frac{LT}{TA} \cdot \frac{AB}{BK} \cdot \frac{KS}{SL} = 1 \iff \frac{KS}{SL} = \frac{1}{2} \iff \frac{KL}{SL} = \frac{3}{2} \iff \frac{SL}{KL} = \boxed{\frac{2}{3}}$.