Find the ratio of segments using Ceva's theorem

Question

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 $[AB]$.

Answer 1

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?

Answer 2

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.

Answer 3

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 $\boxed{AX:XB = 5:16}$.