In ΔABC, extend BC to ray BD. Through C, draw CE ∥ BA. Find the exterior angle ∠ACD in terms of ∠A and ∠B.

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