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
Since AD bisects angle BAC
Then let DC= x and BD = 3 -x
So
(3 - x) / 4 = x / 5
5 (3 -x) = 4x
15 - 5x = 4x
15 = 9x
x = 15/9 = 5/3
Height of triangle ADC = 4 base = (5/3)
Area = (1/2) (5/3) 4 = 20 / 6 = 10/3 = 3 (to the nearest integer )