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.
[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 $\lfloor 6 \cdot \frac{1}{2} \rfloor = \boxed{3}.$
To find the area of $\triangle ADC$, we need to calculate the length of side $\overline{AD}$, as well as the height from $D$ to $\overline{AC}$. Then, we can use the formula for the area of a triangle.
From the given information, we know that triangle $ABC$ is a right triangle, with $\angle ABC = 90^\circ$. Let's label the intersection of $\overline{AD}$ and $\overline{BC}$ as point $E$.
Since $\overline{AD}$ bisects $\angle BAC$, we have $\angle CAD = \angle DAB = \angle BAD$. This means that triangle $ACD$ is an isosceles triangle, with $AC = AD$.
Using the given information, $AB = 4$ and $AC = 5$. In a right triangle, the lengths of the sides satisfy the Pythagorean theorem, $AB^2 + BC^2 = AC^2$.
Substituting in the given values, we have $4^2 + 3^2 = 5^2$, which simplifies to $16 + 9 = 25$.
So, $25 = 25$, which means the given information is consistent.
Since triangle $ACD$ is isosceles and $AC = AD$, triangle $ACD$ is also a right triangle.
By the Pythagorean theorem, $AC^2 = AD^2 + CD^2$. Substituting the given values, $5^2 = AD^2 + CD^2$. Since $AD = AC$ and $CD = BC$, we have $5^2 = AC^2 + BC^2$.
Substituting $AC^2 = 25$ and $BC^2 = 3^2 = 9$, we have $25 = 25 + 9$. This equation is not satisfied, so the given information is inconsistent. The problem cannot be solved as stated.
Therefore, the area of $\triangle ADC$ cannot be determined.
To find the area of triangle $\triangle ADC$, we can use the formula $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$. In this case, we need to find the base and height of triangle $\triangle ADC$.
First, let's find the length of $\overline{AD}$, which is the height of the triangle. Since $\overline{AD}$ bisects $\angle BAC$, we can use the Angle Bisector Theorem. The Angle Bisector Theorem states that the ratio of the lengths of the two segments created by an angle bisector is equal to the ratio of the lengths of the two opposite sides of the triangle. In other words, $\frac{BD}{CD} = \frac{AB}{AC}$.
Plugging in the given values, we have $\frac{BD}{CD} = \frac{4}{5}$. Since $BD + CD = BC = 3$, we can set up the equation $\frac{BD}{3-BD} = \frac{4}{5}$. Solving this equation, we find $BD = \frac{12}{7}$ and $CD = \frac{9}{7}$.
Next, let's find the length of $\overline{AC}$, which is the base of the triangle. We are given that $AC = 5$.
Now, we can calculate the area of triangle $\triangle ADC$ using the formula $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$. Plugging in the values we found, we have $\text{Area} = \frac{1}{2} \cdot 5 \cdot \frac{12}{7} = \frac{30}{7}$. Rounding to the nearest integer, the area of triangle $\triangle ADC$ is $\boxed{4}$.