AMC 10 · 2022 · #20
Grade 8 geometry-2dPick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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°.
Extend AD and BE until they meet at G. Since AB ∥ DE, triangles △ GDE ∼ △ GAB by AA.
Extend two lines until they meet; parallel sides force the two triangles to be similar.
8.G.A.5Draw A DiagramThe similarity ratio DE/AB = , so GD/GA = — meaning D is the midpoint of AG.
Half-as-long DE vs AB means D sits exactly halfway between A and G on the slanted line.
7.G.A.1Identify SubproblemsF lies on line B-E-G and AF ⊥ BE, so ∠ AFG = 90° — △ AFG is right-angled with hypotenuse AG.
F lives on the extended line BG, so the perpendicular from A lands at a right angle to the hypotenuse AG.
7.G.B.5Identify SubproblemsBy the median-to-hypotenuse theorem, FD = AG = AD = DG; with rhombus sides AD = CD this gives FD = CD.
Median from the right-angle vertex to the hypotenuse is exactly half the hypotenuse — so F, A, G are equidistant from D.
7.G.A.2Identify SubproblemsSince FD = AD = DG = CD, the points F, A, G, C lie on a circle centered at D.
Four points at the same distance from D — that's exactly the definition of a circle through them.
7.G.B.4Identify SubproblemsInscribed angle ∠ GFC = ∠ GDC; with ∠ GDC = 180° - 46° = 134°, so ∠ GFC = 67°.
Inscribed angle is half the central angle on the same arc; supplementary angles fix the central one.
7.G.B.5Identify SubproblemsB, F, G are collinear, so ∠ BFC is the supplement of ∠ GFC: ∠ BFC = 180° - 67° = 113°. Answer (D).
Straight line through B, F, G — the two angles on either side at F sum to 180°.
7.G.B.5Work BackwardsExtend 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 = (134°) = 67°, and the supplement gives ∠ BFC = 113°, choice (D).