🇺🇸 AMC 8 ⇄ switch contest
2000 AMC 8

Problem 24

Problem 24 · 2000 AMC 8 Stretch
Geometry & Measurement angle-chaseexterior-angle
Figure for AMC 8 2000 Problem 24
Show answer
Answer: D — 80°.
Show hints
Hint 1 of 2
The figure looks like a tangle of lines, but the only thing you're TOLD is ∠AFG = ∠AGF — that quietly says triangle AFG is isosceles, and you know its apex ∠A = 20°. Start there; it hands you the angle at F for free.
Still stuck? Show hint 2 →
Hint 2 of 2
Once you have the angle at F, notice that same angle is the *exterior* angle of triangle BFD at F. The exterior-angle theorem says it equals the two far-off angles added together — and those happen to be exactly ∠B and ∠D, the thing you want.
Show solution
Approach: isosceles triangle to get the angle at F, then exterior-angle theorem
  1. ∠AFG = ∠AGF makes triangle AFG isosceles. With apex ∠A = 20°, each base angle is (180° − 20°) / 2 = 80°, so ∠AFG = 80°.
  2. That 80° angle at F is the exterior angle of triangle BFD. The exterior-angle theorem: an exterior angle equals the sum of the two remote interior angles — here those remote angles are exactly ∠B and ∠D.
  3. So ∠B + ∠D = 80°, no need to find B and D separately.
  4. The power move: the exterior-angle theorem lets you grab a SUM of two angles in one shot — perfect when a problem asks for ∠B + ∠D rather than either one alone. Always read whether you're asked for a single angle or a sum; a sum often means 'don't solve them individually.'
Another way — supplement, then angle sum (as MAA presents it):
  1. From the isosceles triangle, ∠AFG = 80°, so the angle inside triangle BFD at F is its supplement, 180° − 80° = 100°.
  2. The angles of triangle BFD sum to 180°, so ∠B + ∠D = 180° − 100° = 80°.
Mark: · log in to save
↑