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Question
In a triangle ABC, if the bisectors of angles ABC and ACB meet at M then prove that: ∠BMC = 90° + `(1)/(2)` ∠A.
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Solution
Since BM and CM are bisectors of ∠ABC and ∠ACB,
∠B = 2∠OBC and ∠C = 2∠OCB ....(i)
In ΔABC,
∠A + ∠B + ∠C = 180°
⇒ ∠A + 2∠OBC + 2∠OCB = 180° ....[From (i)]
⇒ `(∠"A")/(2) + ∠"OBC" + ∠"OCB"` = 90° ....[Dividing both sides by 2]
⇒ ∠OBC + ∠OCB = 90° - `(∠"A")/(2)` ....(ii)
Now, in ΔBMC,
∠OBC + ∠OCB + ∠BMC = 180°
⇒ `90° - (∠"A")/(2) + ∠"BMC"` = 180° ....[From (ii)]
⇒ ∠BMC = `180° - 90° + (∠"A")/(2)`
⇒ ∠BMC = `90° + (∠"A")/(2)`.
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