# Isosceles triangle

Construct an isosceles triangle, given the inradius and one exradius.

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...

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 :

r_{A} = 4R sin(A/2)cos(B/2)cos(C/2) with r_{A} the exradius in angle A.

We deduce at once r/r_{A} = tg(B/2)tg(C/2)

If the triangle is isosceles, we have to considere the two separate cases :

- AB = AC and angles B = C. That is tg²(B/2) = r/r
_{A}, giving the base angle B = C
- 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/r_{A}, hence the base angle A = C