From the problem, we know that $C^{\prime} D^{\prime}$ is symmetric to $C D$ and $C^{\prime} T$ is symmetric to $C T$ with respect to the line $T M$. By the properties of line symmetry, we have $C^{\prime} T = C T$ and $C^{\prime} S = C S$.
Let $A C^{\prime} = x, B T = y$. In the right triangle $\triangle B C^{\prime} T$,
$$
y^{2} + 1 = (x + 1 - y)^{2}
$$

which simplifies to
$$
y = \frac{x^{2} + 2x}{2(x + 1)}
$$

Also, from $\triangle P C^{\prime} S^{\prime} \sim \triangle B T C^{\prime}$, we have $\frac{P C^{\prime}}{B T} = \frac{C^{\prime} S^{\prime}}{C^{\prime} T}$, which means

and thus
$$
\begin{array}{c}
\frac{x - \frac{1}{3}(x + 1)}{y} = \frac{\frac{1}{3}(x + 1)}{x + 1 - y} \\
2x^{2} + x - 3xy = 1
\end{array}
$$

From (1) and (2), we get $x^{3} = 2$, so the length of $A C$ is $\sqrt[3]{2}$.
Example 5 As shown in Figure 1.7.5, let $A B C D$ be a square cardboard, which is divided into three rectangles by lines $P Q$ and $R S$ parallel to $B C$. Fold the cardboard so that point $C$ falls on point $C^{\prime}$ on $A B$, and point $S$ falls on point $S^{\prime}$ on $P Q$, with $B C^{\prime}=1$. Try to find the length of $A C^{\prime}$.