In the diagram, $\overline{CX}$ bisects $\angle ACB$. Find the ratio of the area of triangle $BCX$ to the area of triangle $ACX$. Express your answer as a common fraction.
Triangle $\triangle ACX$ is divided into two right triangles: $\triangle ACB$ and $\triangle BCX$, so the ratio of the area of $\triangle BCX$ to the area of $\triangle ACX$ is the same as the ratio of the leg of the right triangle $\triangle BCX$ to the hypotenuse of the right triangle $\triangle ACX$. Therefore, the ratio of the areas is $\boxed{\frac12}$.
Since $\overline{CX}$ bisects $\angle ACB$, we know that $\angle ACX \cong \angle BCX$. Thus, $\triangle ACX$ and $\triangle BCX$ share an angle and are both right triangles. We can thus find the ratio of their areas by comparing the lengths of their legs.
Let $a$ represent the length of $\overline{AX}$ and $c$ represent the length of $\overline{BX}$. Then $BX=a$ and $AX=c$.
By the Pythagorean Theorem, we have
\[CX = \sqrt{AX^2+AC^2} = \sqrt{c^2+1} \quad \text{and} \quad CX = \sqrt{BX^2+BC^2} = \sqrt{a^2+4}\!.\]Therefore, we have $\sqrt{a^2+4} = \sqrt{c^2+1}$. Squaring both sides gives $a^2+4 = c^2+1$, so $a^2 = c^2-3$.
The area of $\triangle BCX$ is $\frac{1}{2} \cdot BC \cdot BX = \frac{1}{2} \cdot 3 \cdot a = \frac{3a}{2}$.
The area of $\triangle ACX$ is $\frac{1}{2} \cdot AC \cdot AX = \frac{1}{2} \cdot 1 \cdot c = \frac{c}{2}$.
Thus, the ratio of the area of $\triangle BCX$ to the area of $\triangle ACX$ is $\dfrac{\frac{3a}{2}}{\frac{c}{2}} = \boxed{\frac{3a}{c}}$.
To find the ratio of the area of triangle $BCX$ to the area of triangle $ACX$, we need to compare the heights of the two triangles.
Let's call the height of triangle $BCX$ $h_1$, and the height of triangle $ACX$ $h_2$.
Since $\overline{CX}$ is a perpendicular bisector of $\overline{AB}$, we can use the Perpendicular Bisector Theorem to find the ratio of the heights. According to the theorem, the ratio of the lengths of the two segments formed by a perpendicular bisector is equal to the ratio of the areas of the two triangles.
In this case, the ratio of the lengths of $\overline{AX}$ and $\overline{CX}$ is equal to the ratio of the areas of triangle $ACX$ and triangle $BCX$.
So, to find the ratio of the areas, we first need to find the ratio of the heights.
We can use similar triangles to find the ratio of the heights. Triangles $ACX$ and $BXC$ share an angle at point $C$, and since $\overline{CX}$ is a bisector of $\angle ACB$, we know that $\angle ACX$ and $\angle BXC$ are congruent.
By Angle-Angle Similarity, we can conclude that triangles $ACX$ and $BXC$ are similar triangles.
Therefore, the ratio of the heights is equal to the ratio of corresponding side lengths. Since the sides $\overline{BX}$ and $\overline{AX}$ are corresponding side lengths, we have:
$\frac{h_1}{h_2} = \frac{\overline{BX}}{\overline{AX}}$
Now, we can see the lengths of $\overline{BX}$ and $\overline{AX}$ in the diagram, and we can substitute the values into the equation.
After solving for the ratio of heights, we can square it to find the ratio of the areas since the ratio of areas is the square of the ratio of corresponding side lengths.