+1341 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.
+4 Let \(BD=x\). By angle bisector theorem, \(\frac{AB}{BD} = \frac{AC}{CD}\) which is equivalent to \(\frac{4}{x}=\frac{5}{3-x} \Rightarrow 12-4x=5x \Rightarrow 9x=12 \Rightarrow x=\frac{4}{3}\). The area of \([\triangle{ADC}]=\frac{DC\cdot AB}{2}=\frac{(3-\frac{4}{3})\cdot4}{2}=\frac{\frac{5}{3}\cdot4}{2}=\frac{10}{3}\approx3\)
