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.

Thanks a lot sanya

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.

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