# Bisectors of angle B and angle C of a triangle ABC intersect each other at point O.Prove that angle BOC=90 degree+half of angle A.

In a triangle ABC, we have :

angle[A+B+C] = 180 DEGREE [since,sum of the angles of a tri.]

=> 1/2[A+B+C] = 90 DEGREE

=> 1/2A + angle1+angle2 = 90 degree

=> angle1 + angle2 = 90 - 1/2angle A         -------------- equation 1

NOW in trianglel OBC,

angle1 + angle2 + angle BOC  = 180

90 - 1/2 angleA + angle BOC = 180           FROM 1

angle BOC = 90degree + 1/2 angle A.

Hence,it is proved.

• 46

Thanks a lot sanya

• -8

in a triangle ABC we have

angle[A+B+C] = 180 DEGREE [since,sum of the angles of a tri.]

=> 1/2[A+B+C] = 90 DEGREE

=> 1/2A + angle1+angle2 = 90 degree

=> angle1 + angle2 = 90 - 1/2angle A -------------- equation 1

NOW in trianglel OBC,

angle1 + angle2 + angle BOC = 180

90 - 1/2 angleA + angle BOC = 180 FROM 1

angle BOC = 90degree + 1/2 angle A.

Hence,it is proved.

• -7

in triangle ABC

angle A + angle B + angle C = 180 degree ( A.S.P )

angle B + angle C = 180- angle A

1/2 (angle B + angle C) = 1/2 ( 180 - angle A ) = 90 - 1/2 angle A

angle OBC + angle OCB = 1/2 ( 180 - angle A ) = 90 - 1/2 angle A

in triangle OBC

angle BOC = 180 -( angle OBC + angle OCB )

by substituting

angle BOC = 180 - (90 - 1/2 Angle A )

= 90 + 1/2 angle A

hence proved

• 8
Yes totally agree with Sanya
• -10
I dont have ok
• -7
Plzz give the answer properly i cannot understand anything
• -3
We can put angle B as 2x and angle C as 2y • 6
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in triangleABC thebisectors of angleA  and angleB intersect each other at a point O. Then angleA

• 0
In a triangle ABC, we have : angle[A+B+C] = 180 DEGREE [since,sum of the angles of a tri.] => 1/2[A+B+C] = 90 DEGREE => 1/2A + angle1+angle2 = 90 degree => angle1 + angle2 = 90 - 1/2angle A -------------- equation 1 NOW in trianglel OBC, angle1 + angle2 + angle BOC = 180 90 - 1/2 angleA + angle BOC = 180 FROM 1 angle BOC = 90degree + 1/2 angle A. Hence,it is proved.
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