Problem with areas

To find the area of triangle $ACD$, we can use the fact that the ratio of the areas of two triangles with the same altitude is equal to the ratio of their corresponding base lengths.

Let $h$ be the altitude of both triangles $ABC$ and $ADE$ with respect to base $\overline{BE}$. Then, we have:
$$\frac{[ABC]}{[ADE]} = \frac{AB}{AD} = \frac{h}{h} = 1.$$
Since $[ABC] = 4$ and $[ADE] = 8$, we can write the equation:
$$\frac{4}{8} = \frac{AB}{AD}.$$
Simplifying, we find $AB = \frac{1}{2}AD$.

Now, let's focus on triangle $ACD$. We know that $BD = 8$ and $CE = 24$. Since $AB = \frac{1}{2}AD$, we can express $BD$ and $CE$ in terms of $AD$:
$$BD = \frac{1}{2}AB = \frac{1}{4}AD,$$
and
$$CE = 2AB = AD.$$

To find the area $[ACD]$, we need to find the base length $AC$. Using the fact that the ratio of the areas is equal to the ratio of base lengths, we have:
$$\frac{[ACD]}{[ADE]} = \frac{AC}{AD}.$$

Substituting the given values, we have:
$$\frac{[ACD]}{8} = \frac{AC}{AD}.$$

Since $[ADE] = 8$, we can rewrite the equation as:
$$\frac{[ACD]}{8} = \frac{AC}{AD} = \frac{[ACD]}{[ADE]}.$$  Drift Boss

From the above equation, we can conclude that $[ACD] = [ADE] = 8$. Therefore, the area of triangle $ACD$ is $8$ square units.