In ∆ABC , BE perpendicular to AC and CD perpendicular to AB. If BE and CD intersect each other at O and the angle bisectors of ∠OBC and ∠OCB meets at point P, ∠BPC = 148°, then find the measure of ∠A?

In triangle ABC, O is the orthocenter since BE ⟂ AC and CD ⟂ AB. P is the incenter of triangle OBC (intersection of angle bisectors). For any triangle, the angle formed by the intersection of two angle bisectors equals 90° plus half the remaining angle: ∠BPC = 90° + ∠BOC/2. Since quadrilateral AECB is cyclic (both ∠AEB and ∠ACB are right angles), ∠BOC = 180° - ∠A. Substituting: 148° = 90° + (180° - ∠A)/2, which gives ∠A = 64°.