Triangle ex-radius properties

Let's say a triangle ABC has 3 ex-circles.
One of them is touching side BC at D and the extended line segments of AB and AC at E,F.
Then
BD = s - c, CD = s - b and AE = AF = s.
Where AB = c, BC = a and AC = b.

Proof:
And s = semi perimeter = (a + b + c)/2
Why?
By equal tangents property:
AE = AF, BE = BD and CD = CF
Let BE = x => BD = x and CD = a - x = CF
AE = c + x
AF = b + a - x
AE = AF => c + x = b + a - x
=> x = (b + a - c)/2 = s - c
=> BD = x = s - c
=> CD = a - (s - c) = a + c - s = s - b
AE = AF = c + x = s

Next:
Area of the triangle in terms of ex-radius:
Ra = [ABC]/(s-a)
Rb = [ABC]/(s-b)
Rc = [ABC]/(s-c)

Proof:



[ABC] = Area of the triangle ABC.
[ABC] = [DAFIa] - 

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