AMC 8 · 2014 · #9

Grade 8 geometry-2d
angle-sum-triangleisosceles-trianglesupplementary-angles identify-subproblems ↑ Prerequisites: angle-sum-triangle
📏 Medium solution 💡 3 insights 📊 Diagram
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Problem
In △ ABC, point D sits on side AC so that BD = DC, and ∠ BCD = 70°. Find the degree measure of ∠ ADB.

Pick an answer.

(A)
100
(B)
120
(C)
135
(D)
140
(E)
150

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

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.

1STEP 1

△ BDC has BD = DC, so it is isosceles with base BC; its base angles are equal, giving ∠ DBC = ∠ BCD = 70°.

∠ DBC = ∠ BCD = 70°
2STEP 2

Every triangle's interior angles sum to 180°, so ∠ BDC = 180° - 70° - 70° = 40°.

∠ BDC = 180° - 70° - 70° = 40°
3STEP 3

D lies on line AC, so ∠ ADB and ∠ BDC form a straight line and are supplementary: ∠ ADB = 180° - 40° = 140°.

∠ ADB + ∠ BDC = 180° → ∠ ADB = 180° - 40° = 140° → (D)
Answer
140
Check the additivity directly: ∠ ADB + ∠ BDC = 140° + 40° = 180°, which matches the straight line AC at D. Also, 140° is obtuse, which fits the picture — DB leans away from DC (toward A), so the angle on the A-side at D should clearly be the larger of the two pieces. The two base angles of △ BDC are 70° each, and 70° + 70° + 40° = 180° also checks out — every angle in sight adds up correctly.
💡Key takeaway

This 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°.