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
Important Ex Radius Result Number 1:
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:
Important Ex Radius Result Number 2:
Ra = [ABC]/(s-a)
Rb = [ABC]/(s-b)
Rc = [ABC]/(s-c)

Proof:

[ABC] = Area of the triangle ABC.
[ABC] = [CABIa] - [CBIa]
[CABIa] = [CAIa] + [ABIa]
[CAIa] = (base*height)/2 = (AC * DIa)/2 = (AC * Ra)/2
[ABIa] = (base*height)/2 = (AB * FIa)/2 = (AB * Ra)/2
=> [CABIa] = Ra * (AB + AC)/2
[CBIa] = (BC * Ra)/2
=> [ABC] = Ra * (AB + AC - BC)/2 = Ra * (s - a)
=> Ra = [ABC]/(s-a) and similarly the others.

Important Ex Radius Result Number 3:
1/Ra + 1/Rb + 1/Rc = 1/r where is is the inradius.
Proof:
Using the previous result:
1/Ra = (s - a)/[ABC]
1/Rb = (s - b)/[ABC]
1/Rc = (s - c)/[ABC]
Adding all:
1/Ra + 1/Rb + 1/Rc = (3s - (a+b+c))/[ABC] = (3s - 2s)/[ABC] = s/[ABC]
But we know that
Inradius r = [ABC]/s
Hence 1/r = s/[ABC]
Hence proved.