Clearly,

D = 1/2r.AB+ 1 2 r.BC+ 1 2 r.CA,

so that

1 r = AB + BC + CA 2D .

Also,

D = 1/2h1.BC = 1/2h2,     CA = 1/2h3.AB,

thus

1 h1 + 1 h2 + 1 h3 = BC + CA + AB 2D = 1 r .

Now let A¢, B¢ and C¢ be the centres of the escribed circles; see the diagram below. Also, for any triangle with vertices U, V and W, let D(U,V,W) denote its area. "

Considering the area of the kite ABA¢C by splitting it into two triangles in two different ways we get

D(A,B,C)+D(B,A¢,C) = area(ABA¢C) = D(A,B,A¢) + D(A,C,A¢).

This gives (with D = D(A,B,C))

D+1/2r1.BC = 1/2r1.AB + 1/2r1.CA,

and hence

2D r1 = AB + CA - BC.

Similarly,

2D r2 = AB + BC - CA,     2D r3 = CA + BC - AB,

and adding these we get

2D æ ç è 1 r1 + 1 r2 + 1 rC ö ÷ ø = (AB + CA - BC) + (AB + BC - CA) + (CA + BC - AB) = (AB + BC + CA) = 2D r

as required.