Philip's method uses the tangent formula and Sue Liu of Madras College, St Andrew's sent a solution to this problem which depends on the use of Heron's Formula for the area of a triangle. Well done both of you.
triangle
FIRST METHOD
Using the tangents of the angles at the centre of the circle and the formula

tanX = -tan(A + B) = tanA + tanB
tanA tanB - 1

where A + B + X = 180o,
x
r
 =
a
r
 + b
r
ab
r2
 -1
 = r(a + b)
ab - r2
 .
The area pf the triangle is
(a + b + x)r = (a + b)r + r3(a + b)
ab - r2
 = (a + b)abr
ab - r2
as required.

SECOND METHOD
Here is Sue's method. The incircle divides the sides of the triangle into lengths c¢ = a + b, b¢ = b + x and a¢ = x + a as shown in the diagram. The semi-perimeter of the triangle is given by s = x + a + b and from Heron's formula the area of the triangle is
A
=
  ____________________
Ö(s(s - a¢)(s - b¢)(s - c¢))
 
=
  ____________
Ö((a + b + x)abx)
 

Also, the triangle is divided into three smaller triangles and the total area is given by
A
=
1
2
 (a + b)r + 1
2
 (b + x)r + 1
2
 (x + a)r
=
(a + b + x)r .

Equating the two answers


  ___________
Öabx(a + b + x)
 
=
(a + b + x)r
abx(a + b + x)
=
(a + b + x)2 r2
abx
=
(a + b + x)r2
x
=
(a + b)r2
(ab - r2).
Hence
A
=
(a + b + x)r
=
r((a + b)(ab - r2) + (a + b)r2)
ab - r2
=
abr(a + b)
ab - r2.
An alternative method, not using Heron's formula, is based on finding x in terms of a and b using the tangents of the angles at the centre of the circle.