🇺🇸 AMC 8 ⇄ switch contest
1994 AJHSME

Problem 7

Problem 7 · 1994 AJHSME Hard
Geometry & Measurement exterior-angleangle-chase
Figure for AJHSME 1994 Problem 7
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Answer: B — 50°.
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Hint 1 of 2
The angle you want, ∠BDC, lives in triangle BDC, where you already know ∠C = 30°. So the whole job is just finding ONE more angle of that triangle: ∠DBC.
Still stuck? Show hint 2 →
Hint 2 of 2
Look at line E–D–B: it's straight, so the angle that line makes with the baseline at B is shared by both triangles. Chase the angle at B over from the left triangle to the right one.
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Approach: angle sum, then a supplement
  1. Left triangle ABE: angles are 60° at A, 40° at E, so the third angle ∠ABE = 180° − 60° − 40° = 80°.
  2. A, B, C sit on one straight line, so ∠DBC and ∠ABE are supplements along that line: ∠DBC = 180° − 80° = 100°. (Ray BD is ray BE since E, D, B are collinear.)
  3. Right triangle BDC: ∠BDC = 180° − 100° − 30° = 50°.
  4. Sanity check: 50° is a small, sharp angle — and ∠BDC does look acute in the picture, so the answer passes the eyeball test.
Another way — exterior-angle shortcut:
  1. ∠DBC is the exterior angle of triangle ABE at B (the base AB extended to C). An exterior angle equals the sum of the two FAR-AWAY interior angles, so ∠DBC = ∠A + ∠E = 60° + 40° = 100° in one step — no need to first find the 80°.
  2. Then triangle BDC gives ∠BDC = 180° − 100° − 30° = 50°. The exterior-angle rule (an outside angle = the two opposite inside angles) is a huge time-saver in angle chases.
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