Question

If the exterior angle bisects of triangle of ABC meet at point O. Prove that angle boc=90°-angle
a/2​

Answer 1

The external angle of angle b will be 180-b

Similarly, external angle of angle c will be 180- c

Now, BO is the angle bisector of external angle of b , so the 2 halves will be 90- b/2 each

Similarly for angle c, the 2 halves will be 90- c/2 each.

Now, angle BOC will be 180-(90-b/2 ) -(90-c/2) as the total sum of angles of a triangle is 180 degrees.

This will give you (b+c)/2. (I)

Now, consider the original triangle ABC, there angle a+b+c =180 degrees

b+c =180- a

put this in (I)

you get

(180- a)/2

=90-a/2

Hence result.

Hope this helps!