Competition · AMC preparation · step 4 of 4

AMC 8 · 2018 · #20

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Parallel-Cut Area Ratio
similar-trianglesarea-trianglesratio-proportion area-differenceidentify-subproblems ↑ Prerequisites: similar-trianglesarea-triangles
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
In △ ABC, point E sits on side AB with AE = 1 and EB = 2. From E we draw DE ∥ BC (so D lands on AC) and EF ∥ AC (so F lands on BC). The four points C, D, E, F form a quadrilateral inside the triangle. What fraction of the whole triangle's area is the area of CDEF?

Pick an answer.

(A)
$\frac{4}{9}$
(B)
$\frac{1}{2}$
(C)
$\frac{5}{9}$
(D)
$\frac{3}{5}$
(E)
$\frac{2}{3}$

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The figure is the heart of the problem, so Tool #1 (Draw a Diagram) comes first: sketch △ ABC, mark E one-third of the way along AB, then draw the two parallels. The picture immediately shows that △ ABC is sliced into three pieces — a small triangle △ ADE near vertex A, another small triangle △ EBF near vertex B, and the leftover quadrilateral CDEF. That decomposition is exactly Tool #7 (Identify Subproblems): instead of attacking [CDEF] head-on, compute the two corner triangles as fractions of [△ ABC] and subtract from 1. Each corner triangle is similar to △ ABC (parallel sides force the same angles), so its area scales as the square of the side ratio — a single idea applied twice.

1STEP 1

Draw and split the triangle

Sketch △ ABC, mark E one-third along AB, draw DE ∥ BC and EF ∥ AC — the triangle splits into three regions: △ ADE, △ EBF, and CDEF.

[△ ABC] = [△ ADE] + [△ EBF] + [CDEF]
2STEP 2

Spot the similar triangles

Because DE ∥ BC and EF ∥ AC, corresponding angles match, so △ ADE ∼ △ ABC with ratio 13\frac{1}{3} and △ EBF ∼ △ ABC with ratio 23\frac{2}{3}.

△ ADE ∼ △ ABC with ratio 1/3, △ EBF ∼ △ ABC with ratio 2/3
3STEP 3

Square the side ratios

Scaling a figure by k multiplies its area by k², so △ ADE covers 19\frac{1}{9} of △ ABC and △ EBF covers 49\frac{4}{9}.

([△ ADE])/([△ ABC]) = (1/3)² = 1/9, ([△ EBF])/([△ ABC]) = (2/3)² = 4/9
4STEP 4

Subtract the two corners

CDEF is everything left over, so its share is 1 − 19\frac{1}{9} − 49\frac{4}{9} = 49\frac{4}{9} of the triangle — choice (A).

[CDEF]/([△ ABC]) = 1 - 1/9 - 4/9 = (9 - 1 - 4)/9 = 4/9 → (A)
Answer
4/9
The two corner pieces use up 19\frac{1}{9} + 49\frac{4}{9} = 59\frac{5}{9} of the triangle, leaving 49\frac{4}{9} for CDEF — a clean fraction that matches choice (A). A sanity check: since E sits closer to A than to B, the corner near B should be the bigger triangle (49\frac{4}{9} vs 19\frac{1}{9}), which is exactly what the picture shows. Also, 49\frac{4}{9} ＜ 12\frac{1}{2}, consistent with the visual that CDEF takes up a bit less than half the triangle.
💡Key takeaway

This AMC 8 problem only needs Grade 8 similar-triangles reasoning — parallel lines make smaller copies whose areas shrink by the square of the side ratio — that you already know!

  • Draw and split the triangle
  • Spot the similar triangles
  • Square the side ratios
  • Subtract the two corners

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