Question

in a triangle ABC the sides AB and AD are produced to E, D. the angle bisector of angle CBE and angleBCD meet at O . then prove that angle BOC=90 '1/2 angle BAC

Answer 1

Ray BO is the bisector of angle CBE

Therefore,angle CBO=1/2of angle CBE

=1/2(180-y)

=90-y/2 (1)

Similarly,ray OC is the bisector of angle BCD

Therefore,angle BCO=1/2 of angle BCD

=1/2(180-z)

=90-z/2 (2)

In triangle BOC,angle BOC+BCO+CBO=180 (3)

Substituting (1,2,3) you get

Angle BOC+90-z/2+90-y/2=180

Angle BOC=z/2+y/2

=1/2(y+z)

But,x+y+z=180 (angle sum property)

y+z=180-x

Angle BOC=1/2(180-x)

=90-x/2

=90-1/2angle BAC