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 $ riangle ADC$. Round your answer to the nearest integer.

Question

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 $	riangle ADC$. Round your answer to the nearest integer.

GPT 3.5 · Answer

[asy]
pair A=(0,0), B=(0,4), C=(3,0), D=(0,8/5);
draw(A--B--C--cycle);
draw(A--D);
label("$A$",A,SW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,NE);
label("$4$",(0,3),W);
label("$3$",(1.5,0),S);
label("$5$",(3/2,2),NW);
[/asy]

Since $\overline{AD}$ bisects $\angle BAC,$ by the Angle Bisector Theorem, $\frac{BD}{DC} = \frac{AB}{AC} = \frac{4}{5}.$ Since $BD + DC = BC = 3,$
\[\frac{4}{5} DC + DC = 3,\]so $\frac{9}{5} DC = 3,$ or $DC = \frac{5}{3}.$ Then $BD = 3 - \frac{5}{3} = \frac{4}{3}.$

The area of triangle $ABC$ is $\frac{1}{2} \cdot 4 \cdot 3 = 6.$ Since the area of triangle $ADC$ is half the area of triangle $ABC,$ the answer is $\lfloor 6 \cdot \frac{1}{2} \rfloor = \boxed{3}.$

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.