Back and my solution follows below. (You might need to make a copy and mark up the dimensions as they appear in the text below)
Circle radii Yellow, Red and Blue are a, b and c respectively
In the diagram, I have drawn a few triangles and highlighted a couple of lines.
The square inside the main triangle has side 4b
The triangle within the square has a hypotenuse = b + c and the lengths of the other two sides are b and 2b – a.
From Pythag (b + a)^2 = b^2 + (2b – a)^2
This expands then contracts to give 3a = 2b
And finally b = 3a/2 (1)
At the top of the main triangle I have drawn three radii each = b
This splits the top triangle (which contains a red circle) into a b x b square and two kite shapes.
The bottom edge of the left hand kite is 3b. And based on symmetry, so is the long sloping top edge of that kite. The short edges of that kite are b
If the long sides of the right-pointing kite are “x” then using Pythag on the top triangle
(3b + x)^2 = (b = x)^2 + (4b)^2
This can be expanded, then simplified to give
4bx = 8b^2
And finally x = 2b
The vertical side of the top triangle is x + b which = 2b
The vertical side of the bottom triangle (which contains the blue circle) is 4b
The top triangle and bottom triangle are “similar” and therefore the ratio of their sides which are 3b/4b ie ¾ must equal the ratio of the radii of their inscribed circles which is b/c
Hence ¾ = b/c or c = 4b/3 (2)
So now I have c in terms of b and b in terms of a
Therefore c in terms of a must be c = 4b/3 = (4/3) (3/2)a = 2a
QED