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.
Since AD bisects BAC
Let BD = 3 - x
Let CD = x
So we have that
BD / AB = CD / AC
(3 - x) / 4 = x / 5
5 (3 - x) = 4x
15 - 5x = 4x
15 = 9x
x = 15/9 = 5/3 = CD
[ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3