In triangle $ABC$, $\angle ABC = 90^\circ$, and $D$ is on side $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC$. If $AB = 4,$ $BC = 3$, and $AC = 5,$ then find the area of $\triangle ADC$. Round your answer to the nearest integer.
A
4 5
B D C
3
Let BD = 3- x and DC = x
By the Angle Bisector Theorem
BD/AB = DC / AC
(3-x) / 4 = x / 5
5(3-x) = 4x
15 - 5x = 4x
15 = 9x
x = 15/9 = 5/3
[ADC ] = (1/2)DC * AB = (1/2)(5/3)(4) = 10 / 3