Dear Student, Let us take M as O Then the diagram is: In triangle ABC, ∠A +∠B +∠C =180°∠B +∠C = 180° - ∠A 12(∠B +∠C) = 90° - ∠A2 In triangle OBC,∠BOC = 180° - (∠OBC +∠OCB) = 180° -12(∠B +∠C) = 180°- 90° + ∠A2so, ∠BOC = 90° + ∠A2 Hope this information will clear your doubts about topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

Please prove it

Dear Student,
Let us take M as O. Then the diagram is:

In triangle ABC,

∠ A + ∠ B + ∠ C = 180 ° ∠ B + ∠ C = 180 ° - ∠ A \frac{1}{2} (∠ B + ∠ C) = 90 ° - \frac{∠ A}{2} I n t r i a n g l e O B C, ∠ B O C = 180 ° - (∠ O B C + ∠ O C B) = 180 ° - \frac{1}{2} (∠ B + ∠ C) = 180 ° - 90 ° + \frac{∠ A}{2} s o, ∠ B O C = 90 ° + \frac{∠ A}{2}

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

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