prove that:

Let the vertical angle be C. So, the base angles are A and B.

the internal bisector of angle A will make two equal angles of measure A/2 each.

The external bisector of the other base angle will make two equal angles of measure (180-B)/2.

Let both bisectors meet at say P.

So, angle PAB=A/2

angle PBA=B+([180-B]/2)=90+(B/2)

So, in triangle PBA,

angle P+ angle PAB+angle PBA= 180 ( angle sum property)

So,

 angle P=180-(angle PAB+angle PBA)

=180-[(A/2)+90+(B/2)]=90 -[(A/2)+(B/2)]

=90-[(A+B)/2]

=90-[(180-C)/2]

=90-[90-(C/2)]

=C/2

=Half of the vertical angle C.