Question

9. In a ABC, BO and CO bisect interior angles B and C respectively. Prove that
BOC = 90°+½ .A.​

Answer 1

The solution is very straight forward:

In Triangle OBC the angles

OBC = 1/2 (BAC + BCA)

OCB = 1/2 (BAC + CBA)

By angle sum property

BOC = 180 - 1/2 (BAC +BAC+ BCA + CBA)

BOC = 180 - 1/2 (BAC + 180)

BOC = 180 - 90 - 1/2 BAC

BOC = 90 - 1/2 BAC

Answer 2

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