Example 2 As shown in Figure 2, the side lengths of square $ABCD$ and square $CGEF$ are $2$ and $3$, respectively, and points $B$, $C$, and $G$ are collinear. $M$ is the midpoint of line segment $AE$, and $MF$ is connected. Then the length of $MF$ is $\qquad$
(2006, Zhejiang Province Junior High School Mathematics Competition Preliminary)
Analysis 1: As shown in Figure 2, since $M$ is the midpoint of $A E$, it leads to the idea of a midline. The other midpoint that comes to mind is the center of symmetry of the square. Therefore, connect $F G$ and $C E$ to intersect at point $O$, then connect $A C$. It is easy to see that $F G \parallel A C$; connect $M O$. Since $M O$ is the midline of $\triangle A C E$, we have $M O \parallel A C$. Thus, $F, M, O, G$ are collinear. Hence,
$$
F M = F O - M O = \frac{1}{2} F G - \frac{1}{2} A C = \frac{\sqrt{2}}{2}.
$$

Analysis 2: As shown in Figure 3, since $M$ is the midpoint of $A E$, it leads to the idea of a midline.
If we consider making point $F$ the midpoint, extend $E F$ to $H$ such that $E F = F H$. Thus, $A H = 2 F M$. The length of $A H$ can be calculated by constructing a right triangle, i.e., draw $A K \perp E H$ at $K$, then $H K = 1, A K = F D = 1$.
Therefore, $A H = \sqrt{2}$.
So, $F M = \frac{\sqrt{2}}{2}$.

Analysis 3: As shown in Figure 4, since $M$ is the midpoint of $A E$, and $A D \parallel E F$,
consider directly rotating $\triangle F M E$ around $M$ by $180^{\circ}$. Thus, the symmetric point of $F$ is $F'$, which is the intersection of $F M$ and the extension of $A D$. Therefore, $F F' = 2 F M$. Since $D F' = A F' - A D = E F - A D = 1, F D = 1$, we have $F F' = \sqrt{2}$. So, $F M = \frac{\sqrt{2}}{2}$.