Triangle ratio of areas This is a photo that was originally posted on Google Plus. I would like to know how to solve for S. I started by splitting S into two parts S1 and S2 by drawing a line from A to M. 
I also know that I should use the fact that since the triangles share the same base, the ratio of areas = ratio of heights. 
What I have so far:
5/15 =
8/18 = 
Not sure what to equate these two equations to. 


Let $x$ be the area of the triangle $AML$, and $y$ of $AMK$. Note that as $MBC$ and $LBC$ have the same bases, their heights must be in the ratio of $2/3$ and so must be the segments
$$\frac{MC}{LC}=\frac{2}{3}\Rightarrow\frac{LM}{LC}=\frac{1}{3}.$$
By a similar argument, we get
$$\frac{MB}{KB}=\frac{10}{18}=\frac{5}{9}\Rightarrow\frac{KM}{KB}=\frac{4}{9}.$$
Now thinking in the segment $AB$ as the base and about the triangles $AMB$ and $ABC$, we have
$$\frac{x+5}{23+x+y}=\frac{ML}{CL}=\frac{1}{3}.$$
Also, thinking of $AC$ as the base, we have
$$\frac{8+y}{23+x+y}=\frac{MK}{BK}=\frac{4}{9}.$$
Solving for $x$ and $y$, yields $x=10$ and $y=12$. So $S=x+y=22$.
