# 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.

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That...is amazing handwriting. –  mathematics2x2life Nov 19 '13 at 23:51

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 $S=x+y=22$.