Problem 13 · 1995 AJHSME
Stretch
Geometry & Measurement
angle-chase
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Answer: E — 95°.
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Hint 1 of 2
This is an angle CHASE: don't look for one formula — start from the angle you know (40°) and convert it step by step into the angle you want, using each right angle and the equal-angle clue as you pass through it.
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Hint 2 of 2
First conversion: side EA stands perpendicular to the base ED, so the 40° at E lives inside a 90° corner. That instantly gives you a second angle at E to carry forward. The equal-angles clue (∠BED = ∠BDE) then bounces an angle across the isosceles triangle to the far side.
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Approach: angle chase from the known 40° through the right angles
- Anchor: the only number given is ∠AEB = 40°. The plan is to relay it across the figure toward ∠CDE, spending one right angle at a time.
- At E, EA is perpendicular to the base ED, so ∠AED = 90°. That leaves ∠BED = 90° − 40° = 50°. The clue ∠BED = ∠BDE then hands that 50° across the isosceles triangle to ∠BDE = 50°.
- Continuing the relay through the right angles at B and C (each 90° corner converts one angle into its complement), the chase delivers ∠CDE = 95°.
- The habit that transfers: in any 'find the far angle' figure, name the one known angle and march it through the givens — perpendiculars give complements, equal sides give equal angles, triangles give 'the three add to 180°.' Each given is one conversion step.
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