Problem 24 · 1996 AJHSME
Hard
Geometry & Measurement
angle-bisectortriangle-sum
The measure of angle ABC is 50°. AD bisects angle BAC, and DC bisects angle BCA. The measure of angle ADC is
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Answer: C — 115°.
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Hint 1 of 2
You don't need the two base angles separately — only their SUM. The three angles of triangle ABC add to 180°, so the two base angles ∠A + ∠C must total 180° − 50°. Keep them lumped together.
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Hint 2 of 2
Inside the little triangle ADC, the two angles at A and C are each HALVED by the bisectors, so they add to half of (∠A + ∠C). Then the triangle-sum 180° finishes ∠ADC.
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Approach: work with the SUM of the base angles, then halve
- In triangle ABC the base angles satisfy ∠BAC + ∠BCA = 180° − 50° = 130°. We never need them individually — just their sum.
- The bisectors split each in half, so in triangle ADC the angles at A and C add to ½ · 130° = 65°. Then ∠ADC = 180° − 65° = 115°.
- Why this transfers: the angle between two internal bisectors is always 90° + ½(third angle) — here 90° + ½·50° = 115°. And the key habit: when only a SUM of angles is needed, halving the sum beats finding each angle.
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