+128408 Let C = (0,0) A = (4,0) and B = (0,3)
We have a 3-4-5 right triangle so BA = 5
We can find the coordinates of the incenter as ( [ 4(3) ] / 12 , [ 3(4) / 12 ] = (1 , 1)
Equation of line through A and I .... y = (-1/3)(x - 4)
When x = 0 D = ( 0, 4/3)
Equation of line through B and I ...... y = (-2 / 1) x + 3
When y = 0, E = (0, 3/2)
DE = sqrt [ (4/3)^2 + (3/2)^2] = sqrt [ 16/9 + 9/4 ] = sqrt [ 145 ] / 6
