Question

In a triangle ABC the internal bisectors of Angle B and angle C meet at O prove that OA is the internal bisector of angle A

Answer 1

Answer:

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)

Answer 2

Answer:

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