Find the ratio of segments using Ceva's theorem I am going through $IB$ further math geometry topic and just learned Ceva's theorem. Below is one of the questions in the exercise. I have thought for quite a while and cannot solve it. Can anyone give me some clues or hints, please? Thanks.
In the diagram, $BZ:ZC=2:1$ and $AR:RS:SZ=5:4:3$. Find the ratio in which $X$ divides $[AB]$. 

 A: The proof of Ceva'sTheorem uses the fact that if $ A, B $ and $ C $ are three non-collinear points then any vector $  P $  can be expressed as $ P = xA + yB + zC $ where $ x+ y + z = 1 $. Also, a point on the line joining $ A $ and $ B $ is given by $ P = tA + (1-t)B $.
$ Z = \frac {2}{3} C + \frac {1}{3} B $ 
$ S = \frac {3}{4} Z + \frac {1}{4} A $ 
$ = \frac {1}{2} C +\frac {1}{4} B + \frac {1}{4} A $
Similarly, $ R = \frac {5}{18} C + \frac {5}{36} B + \frac {7}{12} A $
$ Y = tS + (1-t)B $
$ = \frac {t}{2} C + (\frac {t}{4} + 1-t)B + \frac {t}{4} A $ 
For $ Y $ to lie on $ AC $,  $ t = \frac {4}{3} $ .
Then $ Y = \frac {2}{3} C + \frac {1}{3} A $ 
$ X = kR + (1-k)Y $ 
This gives $ X $ in terms of $ A, B $ and $ C $.
Find $ k $ for which $ X $ lies on $ AB $ etc.
A: Here, for a start, we have triangle $AZC$ and transversal $Y-B-S$:
$$ {AY\over YC} \cdot {CB\over ZB}\cdot{ZS \over SA} =1$$ so $${AY\over YC} = 2$$
Then we have triangle $BCY$ and transversal $A-S-Z$:
$$ {AY\over CA}\cdot {CZ\over ZB}\cdot {BS \over SY} =1$$ so $${BS\over SY} = 3$$
Finally we have triangle $ABS$ and transversal $Y-R-X$. Can you finish? 
A: Here is one more alternative solution with homotheties:
Let $\mathcal{H}_{T,k}$ denote a homothety with center at $T$ and a extension factor $k$.
We have a following fact (Theorem):
If $\mathcal{H}_{M,k_1}$ and $\mathcal{H}_{N,k_2}$ are homotheties then their compostion $\mathcal{H}_{M,k_1}\circ \mathcal{H}_{N,k_2}$ is again some homothety $\mathcal{H}_{K,k}$ with $k=k_1k_2$ (if $k\ne 1$) and it center $K$ lies on a line $MN$.
Since we have:
\begin{align}
\mathcal{H}_{A, {4\over 3}}: & \;S \longmapsto Z\\
\mathcal{H}_{C,{3}}: &\; Z \longmapsto B\\
\end{align}
we see that $Y$ is a center of homothety which takes $S$ to $B$ with ratio $k=4$.
Now again
\begin{align}
\mathcal{H}_{R, {-4\over 5}}: & \;A \longmapsto S\\
\mathcal{H}_{Y,{4}}: &\; S \longmapsto B\\
\end{align}
we see that $X$ is a center of homothety which takes $A$ to $B$ with ratio $k={-16\over 5}$, so $\boxed{AX:XB = 5:16}$.
