In triangle ABC, points D and E lie on sides AB and AC respectively such that DE is parallel to BC and AD : DB = 4 : 5. Segments CD and BE intersect at F. What is the ratio of the area of triangle DEF to the area of triangle CBF?

Introduction / Context:
This geometry problem involves a triangle with a segment parallel to one side, creating similar triangles, and then concurrent segments intersecting at a point inside the triangle. The question asks for the ratio of areas of two smaller triangles formed by these constructions. It tests understanding of similarity, area ratios in similar triangles, and the effect of cevians (segments like BE and CD) on area distribution when they intersect inside a triangle.

Given Data / Assumptions:

• Triangle ABC is a general triangle.
• D is on AB and E is on AC such that DE ∥ BC.
• AD : DB = 4 : 5, so D divides side AB in that ratio.
• Segments CD and BE intersect at F.
• We must find area(ΔDEF) : area(ΔCBF).
• No specific side lengths are given, so we can assume convenient coordinates for the vertices.

Concept / Approach:
Since DE is parallel to BC, triangle ADE is similar to triangle ABC. The ratio AD : AB gives the linear scaling between these triangles. From similarity, the ratio of their areas equals the square of the ratio of corresponding sides. After finding the area of ΔADE in terms of area of ΔABC, we infer how much area is left in the trapezoidal region between DE and BC. Using a coordinate approach, we then locate D and E, find the intersection F of BE and CD, and compute the areas of ΔDEF and ΔCBF to get their ratio. Because the scaling is consistent, the final ratio is independent of the particular coordinates chosen.

Step-by-Step Solution:
Let us place triangle ABC in coordinates for convenience.Take A = (0, 0), B = (9, 0), and C = (0, 9), so AB = 9 and AC = 9, forming a right triangle for easier arithmetic.On AB, AD : DB = 4 : 5, so AB = 9 corresponds to 4k + 5k = 9k, giving k = 1.Thus AD = 4 and DB = 5, so D = (4, 0).DE ∥ BC. The line BC goes from B(9, 0) to C(0, 9) with direction vector (−9, 9).A line through D that is parallel to BC has the same direction, so starting from D(4, 0) and moving in direction (−9, 9) will meet AC.AC runs from A(0, 0) to C(0, 9), so it is the y-axis segment.The line through D has equation x = 4 − 9t, y = 0 + 9t.It meets AC where x = 0, so 4 − 9t = 0 gives t = 4 / 9, so y = 9 * (4 / 9) = 4.Hence E = (0, 4).Now, line BE connects B(9, 0) and E(0, 4), and line CD connects C(0, 9) and D(4, 0). Their intersection is F.By solving these line equations, we obtain F = (36 / 13, 36 / 13).We now compute area(ΔDEF) and area(ΔCBF).Using coordinate geometry or polygon area formulas, area(ΔDEF) = 40 / 13 and area(ΔCBF) = 405 / 26.The ratio area(ΔDEF) : area(ΔCBF) = (40 / 13) : (405 / 26) = (40 / 13) * (26 / 405).Simplify: (40 * 26) / (13 * 405) = (40 * 2) / 405 = 80 / 405 = 16 / 81.Thus area(ΔDEF) : area(ΔCBF) = 16 : 81.

Verification / Alternative check:
We can reason more conceptually. Since AD : AB = 4 : 9, triangle ADE is similar to triangle ABC with linear ratio 4 : 9. Therefore, area(ΔADE) : area(ΔABC) = 4^2 : 9^2 = 16 : 81. The region between DE and BC accounts for the remaining area, which is area(ΔABC) − area(ΔADE) = (81 − 16) / 81 of the total. It is consistent that ΔDEF is a small part of ΔADE and ΔCBF is a part of the lower larger region, giving a ratio of 16 : 81 overall. The coordinate computations align with this conceptual expectation.

Why Other Options Are Wrong:
The ratio 16 : 25 would arise from incorrectly using side length ratios rather than squared side ratios. The ratio 81 : 16 is simply the inverse of the correct answer. The ratios 4 : 9 and 25 : 16 come from misapplied similarity arguments or mixing linear and area ratios. Only 16 : 81 matches both the precise coordinate geometry calculation and the similarity-based reasoning.

Common Pitfalls:
Students often confuse linear similarity ratios with area ratios and may use 4 : 9 directly instead of 16 : 81. Others might misplace points D and E or incorrectly assume symmetric positions that do not match the ratio AD : DB = 4 : 5. Algebraic mistakes while solving for F or computing triangle areas can also occur. Working with clear coordinates and remembering that area scales with the square of the linear factor helps avoid these errors.

Final Answer:
The ratio of the areas of triangles DEF and CBF is 16 : 81.