# 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 $4/3$?

We have $AD=4$ and $\frac{AE}{ED}=\frac 13$, so $4=AD=AE+ED=ED(\frac 13+1)=\frac43ED$, so $ED=3, AE=1$

$AD$ consists of two parts: $AE$ and $DE$. The long part, $DE$ is three times longer than the short part $AE$. Together, therefore, they must be four times longer.

ABE is similar to EDC (not ADC), so AE and ED have the ratio of 1:3 and together they add up to 4 (AD), so AE is 1 and ED is 3.

And thus, to get the area of AEC, we need to take the area of ADC (4*9/2) and subtract the area os EDC (3*9/2), which equals 4.5

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 $$AD=4$$ $$AE=1$$