In triangle $ABC$, $D$ lies on segment $\overline{BC}$ such that $\overline{AD}$ is an angle bisector. If $AB = 2$, $AC = 2$, and $BC = 19$, 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.)
Let $E$ be on $\overline{AB}$ such that $\overline{CE}$ is an angle bisector. Note that triangle $ABC$ has area
\[\frac{1}{2} \cdot 2 \cdot 2 \cdot \sin A = 2 \sin A,\]and triangle $ACE$ has area $\frac{1}{2} \cdot 2 \cdot 2 \cdot \sin \frac{A}{2} \cos \frac{A}{2} = 2 \sin \frac{A}{2} \cos \frac{A}{2} = \sin A.$
[asy]
unitsize (0.5 cm);
pair A, B, C, D, E;
A = (5,4);
B = (0,0);
C = (20,0);
D = interp(B,C,2/3);
E = interp(A,B,3/5);
draw(A--B--C--cycle);
draw(A--D);
draw(A--E);
draw(C--E);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NW);
[/asy]
Then
\[\frac{BD}{CD} = \frac{AB}{AC} \cdot \frac{\sin \angle BDA}{\sin \angle CDA} = \frac{2}{2} \cdot \frac{\sin \angle BDA}{\sin \angle CDA} = \frac{\sin \angle BDA}{\sin \angle CDA}.\]Since triangles $ABD$ and $ACD$ have the same height (namely, $AD$) from bases $BD$ and $CD,$ respectively,
\[\frac{[ABD]}{[ACD]} = \frac{BD}{CD}.\]Thus,
\[\frac{[ABD]}{[ACD]} = \frac{\sin \angle BDA}{\sin \angle CDA} = \frac{\sin \angle BDA}{\sin \angle BEA} = \frac{DB}{EB} = \frac{2}{2} = \boxed{1}.\]
To find the ratio of the area of triangle $ABD$ to the area of triangle $ACD$, we need to find the lengths of $AD$, $BD$, and $CD$.
We can use the Angle Bisector Theorem to find the lengths of $BD$ and $CD$. According to the Angle Bisector Theorem, the ratio of the lengths of the segments that the angle bisector divides the opposite side into is equal to the ratio of the lengths of the two sides of the triangle that form the angle.
Let $BD = x$ and $CD = 19 - x$.
Applying the Angle Bisector Theorem, we have
\[\frac{x}{19-x} = \frac{2}{2} = 1.\]
Simplifying the equation, we get
\[x = 19 - x.\]
Solving for $x$, we find
\[2x = 19,\]
\[x = \frac{19}{2}.\]
Therefore, $BD = \frac{19}{2}$ and $CD = 19 - \frac{19}{2} = \frac{19}{2}$.
Now, let's find the length of $AD$. Using the angle bisector theorem again, we have
\[\frac{AD}{DB} = \frac{AC}{CB}.\]
Substituting the values we know, we get
\[\frac{AD}{\frac{19}{2}} = \frac{2}{19}.\]
Simplifying the equation, we find
\[AD = \frac{4}{19}.\]
Now, we can find the areas of triangles $ABD$ and $ACD$ using the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
The base of triangle $ABD$ is $BD = \frac{19}{2}$ and the height is $AD = \frac{4}{19}$. Therefore, the area of triangle $ABD$ is
\[\text{Area}_{ABD} = \frac{1}{2} \times \frac{19}{2} \times \frac{4}{19} = \frac{19}{4}.\]
Similarly, the base of triangle $ACD$ is $CD = \frac{19}{2}$ and the height is $AD = \frac{4}{19}$. Therefore, the area of triangle $ACD$ is
\[\text{Area}_{ACD} = \frac{1}{2} \times \frac{19}{2} \times \frac{4}{19} = \frac{19}{4}.\]
Finally, we can find the ratio of the area of triangle $ABD$ to the area of triangle $ACD$:
\[\frac{\text{Area}_{ABD}}{\text{Area}_{ACD}} = \frac{\frac{19}{4}}{\frac{19}{4}} = \boxed{1}.\]
To find the ratio of the areas of triangles $ABD$ and $ACD$, we need to find the lengths of the sides $BD$ and $CD$.
Since $AD$ is the angle bisector of angle $A$, we can apply the angle bisector theorem to find the ratio of $BD$ to $CD$. 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 sides opposite those segments.
Applying the angle bisector theorem, we have:
$\frac{BD}{CD} = \frac{AB}{AC}$
Plugging in the given values, we get:
$\frac{BD}{CD} = \frac{2}{2}$
Simplifying, we find that $BD = CD$.
Since triangles $ABD$ and $ACD$ share a common height (altitude) from vertex $A$, the ratio of their areas is equal to the ratio of their bases. In this case, the bases are $BD$ and $CD$, which are equal.
Therefore, the ratio of the areas of triangles $ABD$ and $ACD$ is $BD : CD = 1 : 1$.
Hence, the ratio of the area of triangle $ABD$ to the area of triangle $ACD$ is $\boxed{\frac{1}{1}}$.