Let ABC be a triangle and Aa a point on BC between B,C.
h_a : the A-altitude of ABC
r_a : the A-exradius of ABC
r_ab, r_ac : the A-exradii of AAaB and AAaC.
h_a = 2r_ab.r_ac / (r_a - (r_ab + r_ac)) (1)
Now, assume that r_ab = r_ac and r_a = 2h_a.
Denote r_ab / h_a := x
1 = 2x^2 / (2 - 2x) ==> x^2 + x - 1 = 0
0 < x = (- 1 + sqrt5)/2 = 1/φ, where φ is the "golden number" φ = (1 + sqrt5)/2.
Let Iab, Iac be the A-excenters of AAbB, AAcC, resp., and A* the intersection of the line IabIac and the line of AA' (of the A-altitude).
A' divides ΑA* in golden ratio.
If the triangle ABC is isosceles AB = AC, then the foot A' of the A-altitude coincides with the point of contact of the A-excircle and BC:
Let (A),(B) be two externally tangent circles at C with radius of (A) = 4.radius of (B). The line ACB intersects again the circle (B) at D. A tangent from D to circle (A) intersects the common internal tangent of (A),(B) at E. The bisector of the right angle ECA intersects EA at F. Let G be the orthogonal projection of F on AB. C divides GD in Golden Ratio.