In the figure, what is the area of triangle $ABD$? Express your answer as a common fraction.

[asy]
draw((3,0)--(0,0)--(0,4)--(7,0)--(3,0)--(0,4));
draw((0,2)--(7,0));
draw((.5,0)--(.5,.5)--(0,.5));
label("$A$",(0,4),W);
label("2",(0,3),W);
label("$E$",(0,2),W);
label("2",(0,1),W);
label("$C$",(0,0),SW);
label("3",(1.5,0),S);
label("$F$",(3,0),S);
label("4",(5,0),S);
label("$B$",(7,0),SE);
label("$D$",(21/11,16/11),NE);
[/asy]

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Triangle $AEB$ and triangle $EDB$ are similar since they share angle $B$. Therefore, we can use their side lengths to find the length of $BD$.

Using the fact that $\frac{AD}{DB}=\frac{AE}{EB}$, we can set up the equation $\frac{3}{DB}=\frac{2}{4}$, and solve for $DB$, or the denominator of the fraction. Solving, we find that $DB=6$.

Similarly, using the fact that $\frac{BD}{DE}=\frac{2}{4}$, we can set up the equation $\frac{6}{DE}=\frac{2}{4}$, and solve for $DE$, or the numerator of the fraction. Solving, we find that $DE=3$.

Using the formula for the area of a triangle, $\frac{1}{2}$ base times height, we can find the area of triangle $ABD$. The base $AB$ is equal to 7, and the height $DE$ is equal to 3, so the area of triangle $ABD$ is $\frac{1}{2} \cdot 7 \cdot 3 = \boxed{\frac{21}{2}}$.

To find the area of triangle $ABD$, we can use the formula for the area of a triangle: $\text{Area} = \frac12 \times \text{base} \times \text{height}$.

In this case, the base of triangle $ABD$ is $AD$ and the height is the perpendicular distance from point $B$ to line $AD$.

We can find the length of $AD$ using the Pythagorean theorem: $AD = \sqrt{AB^2 + BD^2}$.
$AB = AC + CB = 2 + 3 = 5$, and $BD = BF + FD = 4 + 2 = 6$.
So, $AD = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}$.

To find the height, we need to find the length of the perpendicular line segment from point $B$ to line $AD$.
Since triangle $ABD$ is a right triangle, we can use the formula for the area of a triangle: $\text{Area} = \frac12 \times \text{base} \times \text{height}$.

The base of this triangle is $AD$ and the height is $BF$.

Using the Pythagorean theorem, we find $BF = \sqrt{BD^2 - FD^2} = \sqrt{6^2 - 2^2} = \sqrt{36 - 4} = \sqrt{32}$.

Therefore, the area of triangle $ABD$ is $\frac12 \times AD \times BF = \frac12 \times \sqrt{61} \times \sqrt{32} = \frac12 \sqrt{61 \cdot 32}$.

Simplifying the square root, we have $\frac12 \sqrt{1952}$.

The largest perfect square that divides evenly into $1952$ is $64$, since $64 \cdot 30 = 1920$, and $1952 - 1920 = 32$.

So, we can simplify $\sqrt{1952}$ as $\sqrt{64 \cdot 30}$, which becomes $8\sqrt{30}$.

Therefore, the area of triangle $ABD$ is $\frac12 \cdot 8\sqrt{30} = \boxed{4\sqrt{30}}$.

To find the area of triangle $ABD$, we can use the formula for the area of a triangle: $A = \frac{1}{2} \times \text{base} \times \text{height}$.

In this case, the base is $AD$ and the height is the perpendicular distance from $D$ to the line $AB$.

First, let's find the length of $AD$. We can see that $AD$ is the hypotenuse of the right triangle $ACD$. So we can use the Pythagorean theorem to find its length:
\[AD^2 = AC^2 + CD^2\]
\[AD^2 = 3^2 + 4^2\]
\[AD^2 = 9 + 16\]
\[AD^2 = 25\]
\[AD = 5\]

Now, let's find the perpendicular distance from $D$ to the line $AB$. We can draw a perpendicular line segment from $D$ to $AB$, and let's call the point where it intersects $AB$ as point $G$. Now we have a right triangle $BDG$. We can use the Pythagorean theorem to find the length of $DG$:
\[DG^2 = BG^2 - BD^2\]
\[DG^2 = 2^2 - 5^2\]
\[DG^2 = 4 - 25\]
\[DG^2 = -21\]

But wait! We got a negative value inside the square root, which is not possible for a length. This means that the point $G$ does not exist on segment $AB$.

Since the perpendicular distance $DG$ does not exist, the area of triangle $ABD$ is $\boxed{0}$.

No figure. Cannot draw on these posts.