Competition · AMC preparation · step 4 of 4

AMC 10 · 2022B · #20

Grade 8 geometry-2d
similar-trianglesangle-sum-triangleisosceles-trianglesupplementary-angles identify-subproblemswork-backwards ↑ Prerequisites: similar-trianglesangle-sum-triangle
📏 Long solution 💡 3 insights
Problem
In rhombus ABCD with ∠ ADC = 46°, let E be the midpoint of CD. Let F on BE satisfy AF ⊥ BE. Find ∠ BFC.

Pick an answer.

(A)
110
(B)
111
(C)
112
(D)
113
(E)
114

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Diagram): Sketch the rhombus, mark E on CD, and extend line AD until it meets line BE at a point G. The extension is the unlock — once G exists, two parallel-line triangles (△ GDE ∼ △ GAB at ratio 1:2) hand us D as the midpoint of AG. Tool #7 (Subproblems): break the answer into (i) prove FD = AD via the median-to-hypotenuse theorem in right △ AFG, (ii) show F, A, G, C lie on a circle centered at D, (iii) use the inscribed-angle theorem to convert ∠ GFC from a central angle. Tool #11 (Work Backwards): the target ∠ BFC is supplementary to ∠ GFC (since B, F, G are collinear), so we work backwards from ∠ ADC = 46° to ∠ GDC = 134° to ∠ GFC = 67° to ∠ BFC = 113°.

1STEP 1

Extend the two rays

Extend AD and BE until they meet at G. Since AB ∥ DE, triangles △ GDE ∼ △ GAB by AA.

△ GDE ∼ △ GAB
2STEP 2

Use the similarity ratio

The similarity ratio DE/AB = 12\frac{1}{2}, so GD/GA = 12\frac{1}{2} — meaning D is the midpoint of AG.

GD/GA = 1/2 → AD = DG
3STEP 3

Spot the right angle

F lies on line B-E-G and AF ⊥ BE, so ∠ AFG = 90° — △ AFG is right-angled with hypotenuse AG.

∠ AFG = 90°, AG hypotenuse
4STEP 4

Use the median to the hypotenuse

By the median-to-hypotenuse theorem, FD = 12\frac{1}{2} AG = AD = DG; with rhombus sides AD = CD this gives FD = CD.

FD = AD = DG
5STEP 5

Find the circle centered at D

Since FD = AD = DG = CD, the points F, A, G, C lie on a circle centered at D.

F, A, G, C on circle centered at D
6STEP 6

Apply the inscribed angle theorem

Inscribed angle ∠ GFC = 12\frac{1}{2} ∠ GDC; with ∠ GDC = 180° - 46° = 134°, so ∠ GFC = 67°.

∠ GFC = 1/2 ∠ GDC = 1/2(134°) = 67°
7STEP 7

Take the supplement

B, F, G are collinear, so ∠ BFC is the supplement of ∠ GFC: ∠ BFC = 180° - 67° = 113°. Answer (D).

∠ BFC = 180° - 67° = 113°
Answer
113
Sanity: ∠ ADC = 46°, and the answer 113° = 90° + 23° = 90° + 46°2\frac{46°}{2}. That clean relation ∠ BFC = 90° + 12\frac{1}{2} ∠ ADC matches the inscribed-angle / supplementary-angle chain: 180° - 12\frac{1}{2}(180° - 46°) = 180° - 90° + 23° = 113°. The choices are 110, 111, 112, 113, 114 — only 113 fits the half-angle formula, and the others tempt off-by-one slips in the supplementary step.
💡Key takeaway

Extend AD and BE to meet at G. Then D is the midpoint of the hypotenuse of right △ AFG, so FD = AD = CD — four points F, A, G, C on a circle centered at D. The inscribed angle is half the central one, ∠ GFC = 12\frac{1}{2}(134°) = 67°, and the supplement gives ∠ BFC = 113°, choice (D).

  • Extend the two rays
  • Use the similarity ratio
  • Spot the right angle
  • Use the median to the hypotenuse
  • Find the circle centered at D
  • Apply the inscribed angle theorem
  • Take the supplement

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