Geometry

By the angle bisector theorem, we have AD/DB = AC/BC, or AD/(4) = (AC)/(BC). Similarly, we have BE/EC = AB/AC and CF/AF = BC/AB. Multiplying these equations, we get:

(Area of triangle ABC)^2 = AD*BE*CF*(AC/BC)*(AB/AC)*(BC/AB)
= AD*BE*CF

Now, we can use the fact that the sum of the lengths of the angle bisectors is equal to the perimeter of the triangle, or AI+BI+CI = AB+BC+CA, to find the lengths of the other two angle bisectors. We have:

AI = DI+8 = 11
CI/CA = BI/BA = 8/(AC+8)
CI = (AC*8)/(AC+8)
BI/BA = 8/(AB+8)
BA = (8AB)/(AB+8)

Now, we can use the fact that the area of a triangle can be expressed as (1/2) * base * height, where the height is the length of the altitude to that base. Let h be the length of the altitude from A to BC, and let x be the length of the altitude from D to BC. Then we have:

AD/x = AC/h
x/h = AD/AC = 3/(AC+4)

Similarly, we have:

BE/y = AB/h
y/h = BE/AB = 4/(AC+4)

CF/z = BC/h
z/h = CF/BC = CI/(BC+BA+CI)
z/h = 11/(AC+AB+11)

Now, we can express the area of triangle ABC in terms of x, y, and z:

(Area of triangle ABC) = (1/2) * BC * h
= (1/2) * BC * (x+y+z)
= (1/2) * BC * [(3/(AC+4))x + (4/(AC+4))y + (11/(AC+AB+11))z]

Substituting the expressions for x, y, and z in terms of AC and AB, we get:

(Area of triangle ABC) = (1/2) * BC * [(3AD/(AC+4))BE + (4BE/(AC+4))AD + (11CI/(AC+AB+11))CF]
= (1/2) * BC * [(3*3/(AC+4))*4 + (4*4/(AC+4))*3 + (11*8/(AB+AC+11))*5]

Substituting the values given in the problem, we get:

(Area of triangle ABC) = (1/2) * BC * [12/(AC+4) + 16/(AC+4) + 40/(AB+AC+11)]
= (1/2) * BC * [28/(AC+4) + 40/(AB+AC+11)]

Now, we can use the fact that the area of a triangle can also be expressed as (1/2) * a * b * sin(C), where a and b are two sides of the triangle and C is the angle between them. Let AB = c, BC = a, and CA = b. Then we have:

(Area of triangle ABC) = (1/2) * a * b * sin(C)
= (1/2) * (a/2) * (b/2) * (2 sin(C))
= (1/2) * (a/2) * (b/2) * (2sqrt((s-a)(s-b)(s-c))/ab)
= sqrt((s-a)(s-b)(s-c))*1/4

where s = (a+b+c)/2 is the semiperimeter of the triangle. Substituting a = BC, b = AC, and c = AB, we get:

(Area of triangle ABC) = sqrt((s-BC)(s-AC)(s-AB))*1/4
= sqrt((6)(2)(2)(2))/4
= sqrt(6)

Therefore, the area of triangle ABC is sqrt(6).