In $\triangle A B C$, $\angle A C B=90^{\circ}, A C=$ $B C, D$ is a point on side $A B$, and the perpendicular bisector of line segment $C D$ intersects sides $A C$ and $B C$ at points $M$ and $N$, respectively. If $A D=a$, $B D=b(a, b$ are given positive numbers), try to find the lengths of $C M$ and $C N$ (expressed in the simplest form involving $a, b$), and determine the range of $\frac{b}{a}$. ${ }^{[1]}$.


Solution: As shown in Figure 1, let
$$
C M = D M = x,
$$
and draw
$$
D P \perp A M \text{ at point } P.
$$
Then,
$$
\begin{aligned}
D P & = \frac{a}{\sqrt{2}} = A P, \\
M P & = \left| \left( \frac{a + b}{\sqrt{2}} - x \right) - \frac{a}{\sqrt{2}} \right| = \left| \frac{b}{\sqrt{2}} - x \right|.
\end{aligned}
$$

From $x^{2} = \left( \frac{a}{\sqrt{2}} \right)^{2} + \left( \frac{b}{\sqrt{2}} - x \right)^{2}$, we solve for
$$
C M = x = \frac{a^{2} + b^{2}}{2 \sqrt{2} b}.
$$

Similarly, we get $C N = y = \frac{a^{2} + b^{2}}{2 \sqrt{2} a}$.
Let $\angle A C D = \theta$, then $\angle B C D = 90^{\circ} - \theta$.
When $B D = B C$, $\theta = 22.5^{\circ}$, point $N$ coincides with $B$;
When $A D = A C$, $\theta = 67.5^{\circ}$, point $M$ coincides with $A$.
Therefore, $22.5^{\circ} \leqslant \theta \leqslant 67.5^{\circ}$.
Draw $D Q \perp B C$ at point $Q$. Thus,
$$
\frac{b}{a} = \frac{S_{\triangle C B D}}{S_{\triangle C A D}} = \frac{\frac{1}{2} B C \cdot D Q}{\frac{1}{2} A C \cdot D P} = \frac{D Q}{D P} = \frac{1}{\tan \theta}.
$$

Hence, $\sqrt{2} - 1 \leqslant \frac{b}{a} \leqslant \sqrt{2} + 1$.