Solution: If AB and AC are extended to E and F, it is obtained from the external angle theorem of triangle, and the angles CBE=∠A+∠ACB, ∠BCF=∠A+∠ABC, so ∠ EBC+∠ BCF = 2. Because Bo and Co share ∠CBE and ∠BCF, ∠ CBO+∠ BO, Co =1/2 (∠ CBE+∠ BCF) = 90+1/2 ∞. So ∠ o =180-(∠ CBO+∠ bco) = 90-1/2 ∠ a in △BCO.