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Asked by Smritishraddha suresh

Oct 2, 2013

Bisectors of angles B and C of a triangle ABC intersect each other at O .

Bisectors  of angles B and C of a triangle ABC intersect each other at the point O .

prove <BOC= 90°+1/2 <A

Answers(2)

Answer

Agam Gupta

Member since Sep 21, 2013


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)

Kishore Kumar

Member since


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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