What's the value of the $\frac{AT}{TC}$ in the triangle below? For reference:

In triangle $ABC$, a circle is drawn by $B$
which is tangent to $AC$ at point $T$. This circle intersects $AB$ and $BC$ at points "$R$" and "$S$" respectively.
If $4RB = 3BS$ and $\overset{\LARGE{\frown}}{RT} = \overset{\LARGE{\frown}}{TS}$, calculate $\frac{AT}{TC}$. (Answer:$\frac{3}{4}$)

My progress: I couldn't "develop" this..
only that $OR=OT=OS = r\\\triangle ORT \cong \triangle OTS\ (S.A.S.)$

(figure without scale)

By the tangent and secant theorem we have:
$$AT^2=AR\cdot AB, \quad CT^2 = CS\cdot BC.$$
Therefore,
$$\frac{AT^2}{CT^2}=\frac{AR\cdot AB}{CS\cdot BC}.$$
On the other hand, we know that $BT$ is bissector of $\angle ABC$, then
$$\frac{AT^2}{CT^2}=\frac{AB^2}{BC^2}.$$
So, we got that
$$\frac{AR}{SC}=\frac{AR+RB}{CS+SB} \, \Longleftrightarrow \, SC\cdot AR + RB\cdot CS =AR\cdot SC +AR\cdot SB \, \Longleftrightarrow \, RB\cdot CS= AR \cdot SB \, \Longleftrightarrow \, \frac{AR}{CS}=\frac{RB}{SB}=3/4.$$
And finally,
$$\frac{AT^2}{CT^2}=\frac{AR^2}{CS^2}=\frac{9}{16}\, \Longrightarrow \frac{AT}{CT}=\frac{3}{4}.$$
