Excenter angle at intersection of external bisectors: In ΔABC, sides AB and AC are produced to P and Q respectively. The bisectors of ∠OBC and ∠QCB (external at B and C) meet at O. Find ∠BOC in terms of ∠A.

Introduction / Context:The intersection of the external angle bisectors at B and C is the A-excenter (center of the A-excircle). A classical result gives the angle between lines from this excenter to B and C in terms of ∠A of the triangle.

Given Data / Assumptions:

• Triangle ABC with internal angles A, B, C.
• O is the intersection of the external bisectors at B and C (the A-excenter).
• We seek ∠BOC.

Concept / Approach:Standard angle-chasing in triangle centers yields: at the incenter I, ∠BIC = 90° + A/2; at the A-excenter I_a, ∠B I_a C = 90° − A/2. The construction here corresponds to the excenter case opposite A.

Step-by-Step Solution (identity):Because O is the A-excenter, the known formula applies: ∠BOC = 90° − (1/2)∠A.This comes from supplement relations between internal and external angles and the angle-bisector properties at B and C.

Verification / Alternative check:For an isosceles right triangle with A = 90°, we get ∠BOC = 90° − 45° = 45°, which matches a direct construction.

Why Other Options Are Wrong:The 90° + A/2 formula belongs to the incenter (internal bisectors), not the excenter. 120° ± A/2 does not hold for general triangles.

Common Pitfalls:Confusing incenter and excenter angle formulas; mixing internal vs external bisectors.

Final Answer:90° - 1/2 ∠A