In ΔABC, extend BC to ray BD. Through C, draw CE ∥ BA. Find the exterior angle ∠ACD in terms of ∠A and ∠B.
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A∠A - ∠B
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B1 (∠A + ∠B) 2
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C∠A + ∠B
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D1 (∠A - ∠B) 2
Answer
Correct Answer: ∠A + ∠B
Explanation
Introduction / Context:Exterior angles of a triangle equal the sum of the two opposite interior angles. The introduced parallel line CE ∥ BA helps visualise corresponding angles but the classic exterior-angle theorem already suffices.
Given / Assumptions:
- BC is extended beyond C to ray CD.
- CE ∥ BA.
- We want ∠ACD, the exterior angle at vertex C.
Concept / Approach:The exterior angle at C equals ∠A + ∠B. The parallel line through C confirms that angles corresponding to ∠A and ∠B appear along AC and CD.
Step-by-Step Solution:By the exterior angle theorem: ∠ACD = ∠A + ∠B.With CE ∥ BA, angle chasing shows ∠ACE corresponds to ∠A and angle between CE and CD corresponds to ∠B, summing at C.
Verification / Alternative check:Test with an isosceles or right triangle; numeric measures satisfy the identity.
Why Other Options Are Wrong:Differences or half-sums contradict the exterior-angle theorem.
Common Pitfalls:Mixing up interior angle at C with its exterior; the interior is 180° − (A + B).
Final Answer:∠A + ∠B