Here's the answer for anyone who is curious......
Join AI, BI , CI and DI
The semi-perimeter, S, of triangle ABC = [ 13 + 14 + 15] / 2 = 42/2 = 21
Using Heron's formula to find the the area, we have that the area of ABC
A =√ [ S (S-13) (S-14) (S - 15) ] = √ [21 * 8 * 7 * 6 ] = √ [ 3*7 * 8 * 7 * 3 * 2] =
7*3√ [ 8 *2] = 21√16 = 21 * 4 = 84
Now we have triangles ABI , BCI and CAI with altitudes FI, DI and EI , respectively
And each of these altitudes = the radius of the in-circle
So the area of ABC = S * radius...so we have
84 = 21 * radius
84 / 21 = radius = 4 = altitude of ABI, BCI and CAI
Let BD, BF = x CD, CE = y and AF, AE = z
So..... BD + CD = 14 so x + y =14 (1)
And BF + AF = 13 so x + z = 13 ⇒ -x - z = -13 (2)
And CE + AE = 15 so y + z = 15 (3)
Add (2) and (3) and we have that y - x = 2 add this result to (1) and we have that
2y = 16
y = 8
So z = 7 and x = 6
So the area of AEIF = area of triangle AFI + area of triangle AEI
But triangles AFI and AEI are congruent right triangles....so we can call the area of AEIF = 2 area of triangle AFI
area of triangle AFI = (1/2) ( radius) ( AF) = (1/2) (4) ( z) = (1/2)(4)(7) = 14
So
2* area of triangle AFI = 2 * 14 = 28 = area of AEIF