In ΔABC, sides AB and AC are produced to D and E respectively. BO and CO are the bisectors of ∠CBD and ∠BCE respectively. Then, prove that
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