Question

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Answer 1

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]