Given: A ∆ABC such that bisectors of angle ABC and angle ACB meet at a point O.
To prove: angle BOC = 90°+1/2angle A
Proof: Using angle sum property in ∆BOC, we obtain
angle 1+ angle 2 + angle BOC = 180°...(I)
Using angle sum property in∆ABC, we obtain
angle A + angle B + angle C = 180°
or, angle A + 2(angle 1) + 2(angle 2) = 180° [Since BO and CO are the bisectors of angle ABC and angle ACB respectively, then, angle B = 2(angle 1) and angle C = 2(angle 2)]
or, angle A/2 + angle 1 + angle 2 = 90° [Dividing both sides by 2]
or, angle 1 + angle 2 = 90° - angle A/2 ...(ii)
Substituting this value of angle 1 + angle 2 in (i), we get
90°-angle/2+angle BOC = 180°
or, angle BOC = 180°-90°+angle A/2
or, angle BOC = 90° + 1/2 angle A [Proved]
