Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #24
Grade 7 geometry-2d
Pick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The figure looks busy, but the relevant information lives in just two triangles: AGF (where ∠ A = 20° and the base angles are equal) and BFD (whose interior angles include ∠ B and ∠ D). Tool #7 (Identify Subproblems) splits the figure along these two triangles, linked at point F. Tool #1 (Draw a Diagram) is the sidekick: mark the known 20° at A, label the matching base angles, then use the straight line AFD to carry that information from one triangle to the other. No algebra, just the triangle-angle-sum rule applied twice.
Find the base angle
Subproblem 1: triangle AGF sums to 180° with equal base angles, so each base angle ∠ AFG = 80°.
An isosceles triangle with one 20° vertex angle must have two 80° base angles.
7.G.B.5Identify SubproblemsCarry the angle across F
A, F, D are collinear, so ∠ AFG and ∠ GFD form a linear pair: ∠ GFD = 180° - 80° = 100°.
The straight segment AD acts as a bridge — whatever angle sits on one side of F leaves 180° minus that angle on the other side.
7.G.B.5Draw A DiagramSpot the vertical angle
G lies on ray FB, so rays FG and FB point the same way — thus ∠ BFD and ∠ GFD are the same angle: ∠ BFD = 100°.
Replacing G with B on the same ray doesn't change the angle — angles depend on the ray's direction, not on which point you name on it.
4.G.A.1Identify SubproblemsApply the angle sum
Subproblem 2: the angle sum of triangle BFD gives ∠ B + ∠ D = 180° - 100° = 80°.
Once the third angle of a triangle is pinned down, the sum of the other two is forced — even without knowing each one individually.
In triangle BFD, the two angles ∠ B and ∠ D are forced to add to 80°, even though neither one is fixed on its own.
▸ Why?
The three angles of triangle BFD — ∠ B, ∠ D, and the angle at F — must together make 180°, so ∠ B + ∠ D equals 180° minus the angle at F.
▸ Why?
The triangle's angle at F, namely ∠ BFD, measures 100°, so ∠ B + ∠ D = 180° - 100° = 80°.
▸ Why?
∠ BFD is the same opening as ∠ GFD because B and G both lie on one ray out of F; that opening together with ∠ AFG completes the straight line AFD, so ∠ GFD = 180° - 80° = 100°.
▸ Why?
∠ AFG and ∠ GFD sit side by side along the straight line through A, F, D, and side-by-side angles on a straight line always fill 180°.
▸ Why?
∠ AFG measures 80°, because triangle AGF has apex ∠ A = 20° and two equal base angles, so each base angle is (180° - 20°) / 2 = 80°.
The figure looks tangled, but only two triangles do the work: AGF on top and BFD on the bottom, meeting at F. The isosceles top triangle pins ∠ AFG = 80°; the straight line through F flips that to ∠ BFD = 100°; and the triangle-angle-sum rule in BFD forces ∠ B + ∠ D = 80° — choice (D).
- Find the base angle
- Carry the angle across F
- Spot the vertical angle
- Apply the angle sum
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