In triangle $ABC$, $D$ lies on segment $\overline{BC}$ such that $\overline{AD}$ is an angle bisector. If $AB = 2$, $AC = 2$, and $BC = 3$, then find the ratio of the area of triangle $ABD$ to the area of triangle $ACD$. (Express your answer as a fraction in lowest terms.)
From the Angle Bisector Theorem, we have $\displaystyle\frac{BD}{CD} = \frac{AB}{AC} = 1$. We proceed by dropping an perpendicular from $A$ to $\overline{BC}$, as shown below: [asy]
pair A,B,C,D,EE;
A = (0,0);
B = (1,0);
C= (2.5,0);
EE = (1.5,0);
D = (2,0);
draw(A--B--C--cycle);
draw(A--D);
draw(A--EE);
label("$A$",A,S);
label("$B$",B,S);
label("$D$",D,S);
label("$C$",C,S);
label("$E$",EE,S);
label("$2$",(A+EE)/2,S);
label("$2$",(EE+C)/2,S);
label("$3$",(B+C)/2,S);
[/asy] Therefore, by SAS, we have $\triangle ACD \cong \triangle ABD$, from which it follows that $[ABD] = [ACD]$ and the ratio of the areas of the two triangles is $[ABD]:[ACD] = \boxed{1:1}$.
To find the ratio of the area of triangle $\triangle ABD$ to the area of triangle $\triangle ACD$, we can use the fact that the ratio of the areas of two triangles with the same altitude is equal to the ratio of their bases.
Let $x$ be the length of $\overline{BD}$. Since $\overline{AD}$ is the angle bisector of $\angle BAC$, we have $\frac{BD}{DC} = \frac{AB}{AC}$. Plugging in the given values, we get $\frac{x}{3-x} = \frac{2}{2}$.
Cross-multiplying, we have $x \cdot 2 = (3-x) \cdot 2$, which simplifies to $2x = 6-2x$.
Adding $2x$ to both sides, we get $4x = 6$, and dividing both sides by $4$, we have $x = \frac{3}{2}$.
Thus, $BD = \frac{3}{2}$ and $DC = 3 - BD = 3 - \frac{3}{2} = \frac{3}{2}$.
The area of triangle $\triangle ABD$ is $\frac{1}{2} \cdot AB \cdot BD = \frac{1}{2} \cdot 2 \cdot \frac{3}{2} = \frac{3}{2}$.
The area of triangle $\triangle ACD$ is $\frac{1}{2} \cdot AC \cdot DC = \frac{1}{2} \cdot 2 \cdot \frac{3}{2} = \frac{3}{2}$.
Therefore, the ratio of the area of triangle $\triangle ABD$ to the area of triangle $\triangle ACD$ is $\frac{\frac{3}{2}}{\frac{3}{2}} = \boxed{1}$.
To find the ratio of the area of triangle $ABD$ to the area of triangle $ACD$, we need to compute the areas of both triangles and then take their ratio.
Let's first find the area of triangle $ABD$. We know that $AB = 2$ and $AD$ is an angle bisector. This means that angle $BAD$ and angle $CAD$ are equal, and therefore triangles $ABD$ and $ACD$ share the same height $h$ (the perpendicular distance from $D$ to segment $\overline{AB}$ and segment $\overline{AC}$, respectively).
Let $x$ be the length of segment $\overline{BD}$ and $y$ be the length of segment $\overline{CD}$. Since $\overline{AD}$ is an angle bisector, we can apply the angle bisector theorem to find $\frac{CD}{BD} = \frac{AC}{AB}$. Substituting the given values, we get $\frac{y}{x} = \frac{2}{2} = 1$.
Since $BC = x + y = 3$, we can solve the system of equations $\frac{y}{x} = 1$ and $x + y = 3$ to find $x$ and $y$. From the first equation, $y = x$. Substituting into the second equation, we have $x + x = 3$, so $2x = 3$ and $x = \frac{3}{2}$.
Now we can find the height $h$ using the Pythagorean theorem. In right triangle $ABD$, we have $h^2 + \left(\frac{3}{2}\right)^2 = 2^2$, so $h^2 + \frac{9}{4} = 4$, and $h^2 = \frac{16}{4} - \frac{9}{4} = \frac{7}{4}$. Taking the positive square root, we get $h = \frac{\sqrt{7}}{2}$.
The area of triangle $ABD$ is given by $\frac{1}{2} \cdot \text{base} \cdot \text{height}$, so $[ABD] = \frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{2} = \sqrt{7}$.
Similarly, the area of triangle $ACD$ is also $\sqrt{7}$.
Therefore, the ratio of the area of triangle $ABD$ to the area of triangle $ACD$ is $\frac{\sqrt{7}}{\sqrt{7}} = \boxed{1}$.