# Isosceles triangle

Say in angle A.

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...

Sorry, your browser is not Java compliant.

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.

#### Calculations

Without a prrof right here, mlentio the followin formula for inradius and exradii :

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 :

1. AB = AC and angles B = C. That is tg²(B/2) = r/rA, giving the base angle B = C
2. AB = BC and angles A = C
Therefore B = π - 2C and B/2 = π/2 - C, then tg(B/2) = 1/tg(C) = (1 - tg²(C/2))/ (2 tg(C/2))
We deduce tg²(C/2) = 1 - 2r/rA, hence the base angle A = C