AMC 10 · 2019 · #13

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

inscribed-angleisosceles-triangleangle-sum-trianglearc-measure identify-subproblemscasework ↑ Prerequisites: inscribed-angleisosceles-triangleangle-sum-triangle

📏 Medium solution 💡 3 insights

Problem

△ ABC is isosceles with BC = AC and ∠ ACB = 40^°. A circle is drawn with diameter BC. It meets AC again at D and AB again at E. The diagonals of quadrilateral BCDE are BD and CE; they meet at F. Find ∠ BFC.

Pick an answer.

(A)

(B)

100

(C)

105

(D)

110

(E)

120

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): sketch △ ABC with the circle on diameter BC, and mark the inscribed right angles at D and E (Thales). Tool #7 (Subproblems): peel off three tiny angle-chasing triangles in order — △ ABC gives the base angles; △ BEC (right-angled at E) gives ∠ BCE; △ BDC (right-angled at D) gives ∠ DBC; finally △ BFC adds up to 180^°. Tool #3 matches 110^° to the choices.

1STEP 1

Sketch △ ABC with the circle on diameter BC; mark D on AC, E on AB, and F where diagonals BD and CE cross.

circle on diameter BC; D ∈ AC, E ∈ AB

2STEP 2

Isosceles BC = AC gives ∠ BAC = ∠ ABC, and the angle sum leaves 180^° - 40^° = 140^°, so each base angle is 70^°.

∠ BAC = ∠ ABC =

\frac{180^° - 40^°}{2}

= 70^°

3STEP 3

Thales at E gives ∠ BEC = 90^°, so in △ BEC the third angle ∠ BCE = 180^° - 90^° - 70^° = 20^°.

∠ BEC = 90^°, ∠ BCE = 180^° - 90^° - 70^° = 20^°

4STEP 4

Thales at D gives ∠ BDC = 90^°, so in △ BDC the angle ∠ DBC = 180^° - 90^° - 40^° = 50^°.

∠ BDC = 90^°, ∠ DBC = 180^° - 90^° - 40^° = 50^°

5STEP 5

In △ BFC, ∠ FBC = 50^° and ∠ FCB = 20^°, so ∠ BFC = 180^° - 50^° - 20^° = 110^°.

∠ BFC = 180^° - 50^° - 20^° = 110^°

6STEP 6

Match 110^° to the choices: (D).

110^° → (D)

Answer

110

Since ∠ BFC and ∠ DFE are vertical angles at F, and the other pair of vertical angles is ∠ BFE = ∠ DFC = 70^° (so they sum to 360^° around F), the four angles at F are 110^°, 70^°, 110^°, 70^°. That fits a typical convex quadrilateral with its diagonals crossing inside. 110^° is also obtuse, which matches the picture — F sits a bit below the top vertex A, so ∠ BFC should open wider than 90^°.

💡Key takeaway

This AMC 10 problem only needs Grade 8 angle-chasing you already know! Because BC is a diameter, the angles ∠ BEC and ∠ BDC are right angles (Thales). The isosceles triangle gives base angles 70^°. Inside the two right triangles: ∠ BCE = 20^° and ∠ DBC = 50^°. Then △ BFC closes: 180^° - 50^° - 20^° = 110^°, answer (D).

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