+964 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.
+15000 \(\gamma =atan \frac{4}{3}\\ \gamma =53.1301^{\circ}\\ \alpha=\arcsin \frac{3}{5}\\ \alpha =33.8699^{\circ}\\ \color{blue}\frac{\alpha}{2}=18.4349^{\circ}\)
\(\angle ADC=(180-53.1301-18.4349)^{\circ}\\ \angle ADC=108.4349^{\circ} \)
\(tan \frac{\\alpha}{2}=\frac {\overline{BD}}{4}\\ \overline{BD}=4\cdot tan\ 18.4349^{\circ}\\ \overline{BD}=1\frac{1}{3}\\ \overline{DC}=3-1\frac{1}{3}\\ \overline{DC}=\frac{5}{3}\)
\(A_{ADC}=\frac{\overline{DC} \cdot \overline{AB}}{2}=\frac{5}{3}\cdot \frac{4}{2}=\frac{20}{6}\\ \color{blue}A_{ADC}=3\frac{1}{3}\)
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