Mid-segment parallel to base – area split: In ΔABC, D and E are on AB and AC with DE ∥ BC and AD:DB = 2:3. Find the ratio Area(ΔADE) : Area(quadrilateral BDEC).

Introduction / Context:
When a line segment DE is drawn parallel to BC in triangle ABC, triangles ADE and ABC are similar. The ratio of their areas equals the square of the ratio of corresponding sides (e.g., AD:AB). The remainder of the big triangle is the quadrilateral BDEC.

Given Data / Assumptions:

• DE ∥ BC.
• AD:DB = 2:3 ⇒ AD:AB = 2:(2+3) = 2:5.
• All points are in usual positions (D between A and B; E between A and C).

Concept / Approach:
By similarity, Area(ΔADE)/Area(ΔABC) = (AD/AB)^2 = (2/5)^2 = 4/25. Hence Area(BDEC) = Area(ABC) − Area(ADE) = (25 − 4)/25 = 21/25 of Area(ABC). Therefore, Area(ADE):Area(BDEC) = 4:21.

Step-by-Step Solution:

Verification / Alternative check:
Use coordinates or base-height reasoning; similarity guarantees the squared relation for areas.

Why Other Options Are Wrong:
4:25 and 4:29 treat the partition incorrectly; 4:9 ignores the complementary part’s size.

Common Pitfalls:
Forgetting to subtract from the full triangle’s area when forming the quadrilateral’s area.

Final Answer:
4 : 21