Here's what I come up with :
See the following image :
Let the intersection point of the bisectors = E
Let angle CAB =x angle ABC = y angle ACB = z angle DBC = (180-y)/2
angle EAB = x/2 angle ABE = y+ (180-y)/2
So we have two triangles ABC and AEB
So in ABC .....x + y + z = 180 (1)
And in AEB .....x/2 + 50 + y + (180-y)/2=180 (2)
We can simplify (2) as follows :
x/2 + (180-y)/2 + y = 130 multiply through by 2
x + 180-y + 2y =260
x + y = 80 (3)
Subbing (3) into (1) we have that
(80) + z =180
z = 100 = angle ACB = angle "C"
Which matches your result, CU !!!!