In △ABC, AB is extended to D and AC is extended to E. The bisectors of ∠CBD and ∠BCE meet at O. If ∠A = 40°, find ∠BOC.

Introduction / Context:
Extending AB and AC makes ∠CBD and ∠BCE external angles at B and C. The intersection of these external bisectors is the excenter opposite A. A known result links the angle ∠BOC at this excenter to ∠A of the triangle.

Given Data / Assumptions:

• △ABC with internal angle A = 40°.
• O is the excenter opposite A (intersection of external bisectors at B and C).

Concept / Approach:
For the excenter opposite A, the angle between the lines OB and OC at O equals 90° − A/2. (For the incenter, it would be 90° + A/2.) Here we use the excenter formula.

Step-by-Step Solution:

Verification / Alternative check:
Construct a simple acute triangle (e.g., A=40°, B=70°, C=70°) and verify with angle-chasing that the angle at the A-excenter equals 70°; synthetic geometry texts list this identity directly.

Why Other Options Are Wrong:

• 60°, 65°, 75° correspond to using incorrect ±A/2 adjustments or confusing incenter vs excenter.

Common Pitfalls:
Applying the incenter formula ∠BOC = 90° + A/2 by mistake, or misidentifying external vs internal bisectors.

Final Answer:
70°