CONCURRENT EXTERNAL TANGENTS OF THREE CIRCLES

LEMMA

Let ABC be a triangle and D a point on BC.
Assume that D is between B and C.

Let r_a, r_b, r_c be the radii of the incircles
of the triangles ABC, BAD, CAD, resp.

And R_a, R_b, R_c the radii of the corresponding excircles, ie
Ra := the a-exradius of the triangle BAC
Rb := the a-exradius of the triangle BAD
Rc := the a-exradius of the triangle CAD.

This equality holds:

(r_b / R_b) * (r_c / R_c) = r_a / R_a

Proof 1 (trigonometrically):

In every triangle we have that

Inradius r = 4Rsin(A/2)sin(B/2)sin(C/2)

Exradius r_1 = 4Rsin(A/2)cos(B/2)cos(C/2)

By applying these formulae to the triangles ABC, ABD, ACD, we
get:

r_a / R_a = tan(B/2)tan(C/2)

r_b / R_b = tan(BDA/2)tan(B/2)

r_c / R_c = tan(CDA/2)tan(C/2)

==> (r_b / R_b) * (r_c / R_c) = r_a / R_a

since tan(BDA/2)tan(CDA/2) = 1 (since BDA+CDA = Pi)

Proof 2 (algebraically):

Denote

AB = c, BC = a, CA = b, AD = z, BD = x, DC = y

In every triangle ABC we have that:

Inradius r = area(ABC) / s

Exradius r_1 = area(ABC) / s-a

(where s = semiperimeter of ABC)

Now, by applying these formulae in the triangles
ABC, ABD, ACD, the formula to prove:
(r_b / R_b) * (r_c / R_c) = r_a / R_a

becomes:

(c+z-x)/(c+z+x) * (b+z-y)/(b+z+y) = (-a+b+c) / (a+b+c) ==>

abc + abz + acz + az^2 + axy =

= b^2x + bzx + bcy + byz + bcx + czx + c^2y + cyz (#)

By the Stewart Theorem we get:

b^2x +c^2c = z^2a + axy

The (#) becomes:

abc + abz + acz = bzx + bcy + byz + bcx + czx + cyz

and by replacing x with a-y (since x+y = a), we finally get:

0 = 0

THEOREM

If h_a is the common altitude (from A) of the triangles
ABC, ABD, ACD, then

1. h_a = 2r_b*r_c / (-r_a + r_b + r_c)

2. h_a = 2R_b*R_c / (R_a - R_b - R_c)

In every triangle:

2/h_a = 1/r - 1/r_1 (where r,r_1 are the inradius, a-exradius,resp.)

By applying the formula to the triangles ABC, ABD, ACD,
we get:

2/h_a = 1/r_a - 1/R_a = 1/r_b - 1/R_b = 1/r_c - 1/R_c

From these equalities we get:

R_a = h_a * r_a / (h_a - 2r_a)

R_b = h_a * r_b / (h_a - 2r_b)

R_c = h_a * r_c / (h_a - 2r_c)

By replacing the Ra,Rb,Rc in the

(r_b / R_b) * (r_c / R_c) = r_a / R_a

we get:

h_a = 2r_b*r_c / (-r_a + r_b + r_c)

Similarly, from the equalities above, we get:

r_a = h_a * R_a / (h_a + 2R_a)

r_b = h_a * R_b / (h_a + 2R_b)

r_c = h_a * R_c / (h_a + 2R_c)

and by replacing the r_a,r_b,r_c in the

(r_b / R_b) * (r_c / R_c) = r_a / R_a

we get

h_a = 2R_b*R_c / (R_a - R_b - R_c)