$$\scriptsize(\text{Note: Figure not drawn to scale.})$$

In the right triangle $ABC$ above, $BE=2$ and $AB=12$. If the length of $\overline{AC}$ is 4 units more than 3 times the length of $\overline{DE}$, what is the length of $\overline{DE}$ ?

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Typically, SAT triangle questions involve similarity. In this case, your intuition should tell you that the inner triangle is similar to the overall triangle. We can also prove this because we have two pairs of congruent angles.

$$m \angle B \cong m \angle B \tag*{\tiny reflexive property}$$ $$m \angle BED \cong m\angle BAC \tag*{\tiny right angles}$$

It may be easier to figure out what's going if we decompose the initial figure into the individual triangles.

Since we know that $\triangle ABC \sim \triangle EBD$, we can compare the side lengths to discover the relationship between corresponding sides.

Since $\frac{\overline{BA}}{\overline{BE}}=\frac{12}{2} = 6$, all of the corresponding sides of $\triangle ABC$ should have a length 6 times as large as $\triangle EBD$.

Therefore, $$\overline{AC} = 6 \cdot \overline{DE}$$

The question also gives us the relationship:

$$\overline{AC} = 4+3\overline{DE}$$

Substituting the value of $\overline{AC}$ from one equation into the other:

$$6 \overline{DE} = 4+3\overline{DE}$$ $$3 \overline{DE}=4$$ $$\overline{DE}=\boxed{\frac{4}{3}}$$