In triangle ABC, AB=13,AC=15 , and BC=14. Let be the incenter. The incircle of triangle ABC touches sides BC, AC, and AB at D,

never mind, got it.

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