The easy case is when AB = AC, then the two circles are tangent at the midpoint M of BC and the problem is solved by constructing the common external tangent to the given circles.
More interesting is when AB = BC...
Let S the intersect point of the angle bisector in angle A and the opposite side BC, I and J the incenter and excenter.
Theorem (well known)
ASIJ is an harmonic range (that is IA/IS = - JA/JS)
Projecting this relation on a perpendicular to base allows to
construct the height of vertex A :
A* is the harmonic conjugate of O with respect to I,J*
Hence A lies on a parallel to line BC through A*.
The contact point of the incircle with the base AC is at the midpoint N
of AC, hence on a line midway between line BC and the previous line.
This gives the construction of N (intersection of that line with the given incircle), hence the triangle.
r = 4R sin(A/2)sin(B/2)sin(C/2) with R the circumradius and r the inradius.
and the similar formula :
rA = 4R sin(A/2)cos(B/2)cos(C/2) with rA the exradius in angle A.
We deduce at once r/rA = tg(B/2)tg(C/2)
If the triangle is isosceles, we have to considere the two separate cases :