Question

The sides $AB$ and $AC$ of triangle $ABC$ are poducedc to point $E$ and $D$ respectively. If bisectors $BO$ and $CO$ angle $CBE$ and angle $BCD$ respectively meet at point $O,$ Then prove that
$\angle boc={90}^{\circ }-\frac{1}{2}\angle BAC.$

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