Incenter angle formula: In ΔABC, the angle bisectors of ∠B and ∠C meet at the incenter O. Find ∠BOC in terms of ∠A.

Introduction / Context:The incenter is the intersection of the internal angle bisectors of a triangle. A well-known result gives the angle subtended at the incenter between the lines to vertices B and C.

Given Data / Assumptions:

• Triangle ABC with incenter O.
• We need ∠BOC in terms of ∠A.

Concept / Approach:Classical incenter identity: ∠BOC = 90° + (1/2)∠A. This is derived by partitioning angles at B and C via the bisectors and using the sum of angles in triangle BOC.

Step-by-Step Solution (identity outline):Let ∠B = 2β and ∠C = 2γ (since BO and CO bisect angles).Then ∠BOC = 180° − (β + γ).But β + γ = (B + C)/2 = (180° − A)/2 = 90° − A/2.Thus ∠BOC = 180° − (90° − A/2) = 90° + A/2.

Verification / Alternative check:For an equilateral triangle A = 60°, ∠BOC = 90° + 30° = 120°, which matches symmetry.

Why Other Options Are Wrong:90° − A/2 corresponds to the excenter case; 120° ± A/2 are unrelated to the incenter angle.

Common Pitfalls:Mixing internal and external bisector formulas or mis-halving B and C.

Final Answer:90° + 1/2 ∠A