Question

in a triangle ABC the sides ab and ac are produced to P and Q respectively the bisector of angle PVC and angle Q intersect at a point O prove that angle BOC equal to 90 degree minus 1/2 angle A​

Answer 1

$\bf\large\underline\green{Solution:-}$

➤ Angle BOC = 180 - (Angle CBO + Angle BCO) ----(1)

Since exterior angle of a triangle equal the sum of opposite interior angles, we have:-

Angle CBO = (1/2) angle CBP(Ext. angle) = (1/2) (Angle A + Angle C)

Angle BCO = (1/2) angle BCQ (Ext. angle) = (1/2) (angle A + angle B)

==> angle CBO + angle BCO

= (1/2)[(angle A+angleC) + (angle A + angle B)]

= (1/2)[ (angle A + angle B + angle C) + angle A]

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

From (1) and (2),

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

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