If the bisector of the angle ABC of a triangle ABC meet at point O then prove that angle boc = 90° + 1/2 angle a
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if the bisector of angle ABC and angle ACB of a triangle ABC meet at point O then prove that angle BOC = 90° + 1/2 /_ A
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In ΔABC, by angle sum property we have
2x + 2y + ∠A = 180°
⇒ x + y + (∠A/2) = 90°
⇒ x + y = 90° – (∠A/2) à (1)
In ΔBOC, we have
x + y + ∠BOC = 180°
90° – (∠A/2) + ∠BOC = 180° [From (1)]
∠BOC = 180° – 90° + (∠A/2)
∠BOC = 90° + (∠A/2)
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