In a triangle ABC, a line \(\overline{DE}\) is drawn parallel to \(\overline{BC}\) such that D lies on \(\overline{AB}\) and E lies on \(\overline{AC}\). If \(\overline{AD}:\overline{DB} = 2:3\), find the ratio of the areas of triangle ADE and ABC.
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Key theorem: For similar triangles, \(\text{Area ratio} = (\text{side ratio})^2\). The most common mistake here is confusing \(AD:DB\) with \(AD:AB\). Always find the full side \(AB = AD + DB\) first, then compute the ratio of the partial side to the full side.
Step 1: Set up the similar triangles. Since $DE$ is parallel to $BC$, the angles match up, so triangle $ADE$ is similar to triangle $ABC$. Angle $A$ is shared and the parallel line makes equal corresponding angles.
Step 2: Recall the area rule for similar triangles. For similar triangles, the ratio of areas equals the square of the ratio of matching sides. So we first need the side ratio $AD:AB$.
Step 3: Find $AD:AB$. Given $AD:DB = 2:3$, the whole side $AB = AD + DB = 2 + 3 = 5$ parts. So $AD:AB = 2:5$.
Step 4: Square the side ratio. \[ \left(\frac{AD}{AB}\right)^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25} \]
Step 5: Apply it to the areas. So $\dfrac{\text{area of }ADE}{\text{area of }ABC} = \dfrac{4}{25}$.
Step 6: State the ratio. The areas are in the ratio $4:25$. Therefore \[ \boxed{4:25} \]