Example 5 As shown in Figure 5, given that $\odot O$ is inside rectangle $ABCD$, and tangents to $\odot O$ are drawn through vertices $A, B, C,$ and $D$, touching points are $A_{1}, B_{1}, C_{1}, D_{1}$ respectively. If $AA_{1} = 3, BB_{1} = 4, CC_{1} = 5$, find the length of $DD_{1}$.
Connect $A O$,
$B O$, $C O$, $D O$, $A_{1} O$, $B_{1} O$, $C_{1} O$, $D_{1} O$. Then
$O A_{1} \perp A A_{1}$, $O B_{1} \perp B B_{1}$,
$O C_{1} \perp C C_{1}$, $O D_{1} \perp D D_{1}$.
Let the radius of $\odot O$ be $r$. Then by the Pythagorean theorem, we have
$A O^{2}=A A_{1}^{2}+r^{2}$, $B O^{2}=B B_{1}^{2}+r^{2}$,
$C O^{2}=C C_{1}^{2}+r^{2}$, $D O^{2}=D D_{1}^{2}+r^{2}$.
Draw $O E \perp A B$, $O F \perp D C$, then
$B E=C F$.
Apply the property to Rt $\triangle A O E$ and point $B$, and to Rt $\triangle O D F$ and point $C$ respectively, we get
$$
\begin{array}{l}
A O^{2}=O B^{2}+A B^{2}-2 B A \cdot B E, \\
D O^{2}=O C^{2}+C D^{2}-2 C D \cdot C F .
\end{array}
$$

Subtracting the two equations, we get
$$
A O^{2}-D O^{2}=O B^{2}-O C^{2} \text {, }
$$

which means $A O^{2}+O C^{2}=D B^{2}+O D^{2}$.
Thus, $A A_{1}^{2}+C C_{1}^{2}=B B_{1}^{2}+D D_{1}^{2}$.
Therefore, $D D_{1}=\sqrt{A A_{1}^{2}+C C_{1}^{2}-B B_{1}^{2}}=3 \sqrt{2}$.
[Note] Equation (1) indicates: the sum of the squares of the distances from a point inside a rectangle to two pairs of opposite vertices is equal.