In triangle △ABC, a line DE is drawn parallel to BC such that DE : BC = 3 : 5. What is the ratio of areas ar(△ADE) : ar(trapezium BCED)?

Introduction / Context:
When a segment through a triangle’s sides is drawn parallel to the base, the small triangle near the apex is similar to the whole triangle. Area ratios then come from the square of the linear ratio; the trapezium is the leftover area after removing the small triangle.

Given Data / Assumptions:

• DE ∥ BC.
• DE : BC = 3 : 5 (linear scale).
• We seek ar(△ADE) : ar(BCED).

Concept / Approach:

• Similarity: △ADE ~ △ABC with linear ratio k = 3/5.
• Area ratio: ar(△ADE) : ar(△ABC) = k^2 = (3/5)^2 = 9/25.
• Then ar(BCED) = ar(△ABC) − ar(△ADE) = (1 − 9/25)ar(△ABC) = 16/25 ar(△ABC).

Step-by-Step Solution:

Verification / Alternative check:
Choose coordinates: A(0,0), B(5,0), C(0,5). Line through A parallel to BC at scale 3/5 gives correct ratios, confirming 9:16.

Why Other Options Are Wrong:

• 9:25 confuses triangle-to-whole with triangle-to-trapezium.
• 12:25 and 3:4 do not follow from similar-triangle area scaling.
• None of these: Not applicable since 9:16 is exact.

Common Pitfalls:

• Using linear ratios directly instead of squaring for areas.
• Forgetting to subtract to obtain the trapezium’s area.

Final Answer:
9 : 16