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

Question

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)

Answer 1

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,

$$\frac{AT^2}{CT^2}=\frac{AR^2}{CS^2}=\frac{9}{16}\, \Longrightarrow \frac{AT}{CT}=\frac{3}{4}.$$

Answer 2

Hints:

- Use the power of the points $A$ and $C$ with respect to the circle;

- Use angle bisector theorem for angle $\angle ABC$.

Answer 3

By bisector theorem: $$\frac{m}{n}=\frac{c}{a}$$ $$m^2=cAR=c(c-BR)$$ $$n^2=aCS=a(a-BS)\rightarrow$$ $$\frac{c(c-BR)}{a(a-BS)}=\frac{c^2}{a^2}$$ $$\frac{c-BR}{a-BS}=\frac{c}{a}$$ $$ac-aBR=ac-cBS\rightarrow \frac{c}{a}=\frac{BR}{BS}=\frac{3}{4}$$