<s>[INST] Area of a triangle adjacent to two similar triangles This GMAT problem states:

In the figure above $AD = 4$, $AB = 3$ and $CD = 9$. What is the area of triangle $AEC$
The solution states:

to find the base we need to see that triangles $AEB$ and $CDE$ are similar. The ratio $AB: CD$, is therefore equal to the ratio $AE: ED$. The given information shows that the ratio is $3:9$, or $1:3$. Now dividing $AD$ ($4$) in this ratio gives us $AE$ as $1$.
  The area of $AEC = 1/2\times\text{base} \times \text{height}=1/2 \times 9 = 4.5$

If $AD$ is $4$, and the ratio is $1:3$, how did they get $AE$ to be $1$ and not $4/3$?
 [/INST] Apply Componendo Dividendo on fraction from similar triangles 
$$ \frac{AE}{ED}= \frac{AB}{CD}=\dfrac{3}{9}= \dfrac{1}{3}$$
$$\dfrac{AE}{AE+ED}= \dfrac{1}{1+3}=  \dfrac{1}{4} $$
$$ =\dfrac{AE}{AD}=  \dfrac{1}{4} $$
Since $AD=4$
$$AE=1$$
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