As shown in Figure $3, \odot O$ is inside the rectangle $ABCD$. Tangents are drawn from vertices $A, B, C, D$ to $\odot O$, with points of tangency being $A_{1}, B_{1}, C_{1}, D_{1}$, respectively. If $AA_{1}=3, BB_{1}=4$, $CC_{1}=5$, find the length of $DD_{1}$.
Solution: Let the radius of $\odot O$ be $r$. As shown in Figure 3, connect $O A_{1}, O B_{1}, O C_{1}, O D_{1}$. Then
$$
\begin{array}{l}
O A_{1} \perp A A_{1}, \\
O B_{1} \perp B B_{1}, \\
O C_{1} \perp C C_{1}, \\
O D_{1} \perp D D_{1} .
\end{array}
$$

Connect $A O, B O, C O, D O$. Then $\triangle A O A_{1}$, $\triangle B O B_{1}$, $\triangle C O C_{1}$, $\triangle D O D_{1}$ are all right triangles.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
O A^{2}=A A_{1}^{2}+r^{2}, O B^{2}=B B_{1}^{2}+r^{2}, \\
O C^{2}=C C_{1}^{2}+r^{2}, O D^{2}=D D_{1}^{2}+r^{2} .
\end{array}
$$

Connect $A C, B D$, and let the intersection of $A C$ and $B D$ be $P$, then connect $O P$.
In the rectangle $A B C D$,
$$
A C=B D, A P=P C, B P=P D \text {. }
$$

It is easy to see that $P O$ is the median of $\triangle O A C$ and $\triangle O B D$.
By the median length formula, we have
$$
\begin{array}{l}
O P^{2}=\frac{1}{2} O A^{2}+\frac{1}{2} O C^{2}-\frac{1}{4} A C^{2}, \\
O P^{2}=\frac{1}{2} O B^{2}+\frac{1}{2} O D^{2}-\frac{1}{4} B D^{2} .
\end{array}
$$

Therefore, $O A^{2}+O C^{2}=O B^{2}+O D^{2}$.
Thus, $A A_{1}^{2}+C C_{1}^{2}=B B_{1}^{2}+D D_{1}^{2}$.
Hence $D D_{1}=\sqrt{A A_{1}^{2}+C C_{1}^{2}-B B_{1}^{2}}$
$$
=\sqrt{3^{2}+5^{2}-4^{2}}=3 \sqrt{2} \text {. }
$$
(Bai Yujian, Guo Zhang, Beijing Dongfang Decai School, 100026)