Let $ ABCD$ be a unit square (that is, the labels $ A, B, C, D$ appear in that order around the square). Let $ X$ be a point outside of the square such that the distance from $ X$ to $ AC$ is equal to the distance from $ X$ to $ BD$, and also that $ AX \equal{} \frac {\sqrt {2}}{2}$.  Determine the value of $ CX^2$.
1. **Define the coordinates of the square and point \(X\)**:
   - Let the coordinates of the unit square \(ABCD\) be \(A(0,0)\), \(B(1,0)\), \(C(1,1)\), and \(D(0,1)\).
   - Let \(X\) be a point outside the square such that the distance from \(X\) to \(AC\) is equal to the distance from \(X\) to \(BD\), and \(AX = \frac{\sqrt{2}}{2}\).

2. **Determine the coordinates of \(X\)**:
   - Since \(AX = \frac{\sqrt{2}}{2}\) and \(A(0,0)\), we can place \(X\) at \((-\frac{1}{2}, \frac{1}{2})\) to satisfy the given distance condition.

3. **Verify the distance from \(X\) to \(AC\) and \(BD\)**:
   - The line \(AC\) has the equation \(y = x\).
   - The line \(BD\) has the equation \(y = 1 - x\).
   - The distance from a point \((x_1, y_1)\) to a line \(ax + by + c = 0\) is given by:
     \[
     \text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
     \]
   - For \(AC\): \(a = 1\), \(b = -1\), \(c = 0\):
     \[
     \text{Distance from } X \text{ to } AC = \frac{|(-\frac{1}{2}) - (\frac{1}{2})|}{\sqrt{1^2 + (-1)^2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
     \]
   - For \(BD\): \(a = 1\), \(b = 1\), \(c = -1\):
     \[
     \text{Distance from } X \text{ to } BD = \frac{|(-\frac{1}{2}) + (\frac{1}{2}) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
     \]
   - Both distances are equal, confirming the placement of \(X\).

4. **Calculate \(CX^2\)**:
   - The coordinates of \(C\) are \((1,1)\).
   - The coordinates of \(X\) are \((-\frac{1}{2}, \frac{1}{2})\).
   - Use the distance formula:
     \[
     CX^2 = \left(1 - \left(-\frac{1}{2}\right)\right)^2 + \left(1 - \frac{1}{2}\right)^2 = \left(1 + \frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2
     \]
     \[
     CX^2 = \left(\frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}
     \]

The final answer is \(\boxed{\frac{5}{2}}\).