AMC 10 · 2025 · #8

Grade 11 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
inscribed-anglecyclic-quadrilaterallaw-of-cosinesangle-bisector-theorem identify-subproblems ↑ Prerequisites: inscribed-anglelaw-of-cosines
📏 Medium solution 💡 3 insights
Problem
A five-sided figure ABCDE is drawn with all five corners on one circle. Two angles at corner E are given: angle BEC and angle CED are each 30 degrees. Line AC and line BD cross at point F. Given AB = 9 and AD = 24, find the length BF.

Pick an answer.

(A)
$\frac{57}{11}$
(B)
$\frac{59}{11}$
(C)
$\frac{60}{11}$
(D)
$\frac{61}{11}$
(E)
$\frac{63}{11}$
How to solve
Strategy Draw a Diagram

This is a circle-and-triangle problem, so a clear picture is the anchor: mark the given 30-degree angles and see what they force elsewhere. The picture reveals that triangle ABD holds everything I need, so I split the work into three small pieces (Identify Subproblems): first move the given angles over to vertex A using the inscribed-angle rule, then find side BD, then use where AC hits BD. The middle piece — finding BD from two sides and their included angle — is pure computation, so I turn it into algebra with the Law of Cosines.

1STEP 1

Move the given angles to vertex A

Inscribed angles catching the same arc are equal, so each 30-degree angle at E copies over to vertex A: angle BAC = angle CAD = 30 degrees.

∠ BAC = ∠ BEC = 30°, ∠ CAD = ∠ CED = 30°
2STEP 2

See that AC bisects angle BAD

The two 30-degree halves add up, so angle BAD = 60 degrees and AC bisects it — meaning F is exactly where that bisector meets side BD.

∠ BAD = ∠ BAC + ∠ CAD = 30° + 30° = 60°
3STEP 3

Find BD with the Law of Cosines

Law of Cosines on triangle ABD with sides 9 and 24 and included angle 60 degrees: 81 + 576 - 216 = 441, so BD = 21.

BD² = 9² + 24² - 2(9)(24)cos 60° = 81 + 576 - 216 = 441, BD = 21
4STEP 4

Split BD with the Angle Bisector Theorem

Since AC bisects angle BAD, the Angle Bisector Theorem splits BD in the side ratio: BFFD\frac{BF}{FD} = ABAD\frac{AB}{AD} = 924\frac{9}{24} = 38\frac{3}{8}.

BF/FD = AB/AD = 9/24 = 3/8
5STEP 5

Solve for BF

BF and FD share BD = 21 in the ratio 3 to 8, that is 11 equal parts, so BF is 3 of those parts: BF = 6311\frac{63}{11} — choice (E).

BF = 3/(3+8) · BD = 3/11 · 21 = 63/11
Answer
63/11
Check the split: FD = 21 - 6311\frac{63}{11} = 231−6311\frac{231 - 63}{11} = 16811\frac{168}{11}, so BFFD\frac{BF}{FD} = 63168\frac{63}{168} = 38\frac{3}{8}, which is exactly 924\frac{9}{24}. The ratio holds. BF = 6311\frac{63}{11} is about 5.7, well under half of BD = 21, which fits — the bisector should land closer to B because AB (9) is the shorter of the two sides. Everything is consistent with choice (E).
💡Key takeaway

Equal inscribed angles turn line AC into an angle bisector, so once you know the triangle's third side you just split it in the ratio of the two neighboring sides.

  • Move the given angles to vertex A
  • See that AC bisects angle BAD
  • Find BD with the Law of Cosines
  • Split BD with the Angle Bisector Theorem
  • Solve for BF

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