<s>[INST] 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 $[AB]$. 

 [/INST] 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.
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