Question

bisector of angles B anf C of a triangle ABC intersect each other at point O PROVE THAT ANGLE BOC=90°+1/2ANGLE A​

Answer 1

Answer:

proved

Step-by-step explanation:

In ΔABC,

by angle sum property we have

2x + 2y + ∠A = 180°

⇒ x + y + (∠A/2) = 90°

⇒ x + y = 90° – (∠A/2) ....(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)

proved

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