Here BO, CO are the angle bisectors of ∠DBC &∠ECB intersect each other at O.

∴∠1 = ∠2 and ∠3 = ∠4

Side AB and AC of ΔABC are produced to D and E respectively.

∴ Exterior of ∠DBC = ∠A + ∠C ………… (1)

And Exterior of ∠ECB = ∠A + ∠B ………… (2)

Adding (1) and (2) we get

∠DBC + ∠ECB = 2 ∠A + ∠B + ∠C.

2∠2 + 2∠3 = ∠A + 180°

∠2 + ∠3 = (1 /2)∠A + 90° ………… (3)

But in a ΔBOC = ∠2 + ∠3 + ∠BOC = 180° ………… ( 4)

From eq (3) and (4) we get

(1 /2)∠A + 90° + ∠BOC = 180°

∠BOC = 90° – (1 /2)∠A