AMC 8 · 2014 · #9
Grade 8 geometry-2d
Pick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem is purely geometric, so Tool #1 (Draw a Diagram) is the starting move — sketch △ ABC, mark D on AC, draw BD, and tick the two equal sides BD and DC to make the isosceles structure of △ BDC visible. Once the picture is in place, Tool #7 (Identify Subproblems) splits the angle hunt into two clean steps: (a) find ∠ BDC inside the isosceles triangle △ BDC, then (b) use the fact that ∠ ADB and ∠ BDC form a straight line at D to get ∠ ADB. Each sub-step is a one-line angle calculation.
△ BDC has BD = DC, so it is isosceles with base BC; its base angles are equal, giving ∠ DBC = ∠ BCD = 70°.
Classifying △ BDC as isosceles from its equal side marks is the Grade 5 "classify 2D figures by properties" move.
5.G.B.4Draw A DiagramEvery triangle's interior angles sum to 180°, so ∠ BDC = 180° - 70° - 70° = 40°.
"The three interior angles of a triangle add to 180°" is the Grade 8 informal angle-sum fact.
8.G.A.5Identify SubproblemsD lies on line AC, so ∠ ADB and ∠ BDC form a straight line and are supplementary: ∠ ADB = 180° - 40° = 140°.
Recognizing a linear pair (supplementary adjacent angles on a straight line) is the Grade 7 angle-relationship skill.
7.G.B.5Identify SubproblemsThis AMC 8 problem only needs the Grade 8 fact that a triangle's three angles add to 180° — plus the Grade 7 fact that two angles on a straight line add to 180°.