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.
Let BD=x. By angle bisector theorem, ABBD=ACCD which is equivalent to 4x=53−x⇒12−4x=5x⇒9x=12⇒x=43. The area of [△ADC]=DC⋅AB2=(3−43)⋅42=53⋅42=103≈3