Example 2 (Fill-in-the-blank Question 5) In square $A B C D$, $N$ is the midpoint of $D C$, $M$ is a point on $A D$ different from $D$, and $\angle N M B=$ $\angle M B C$. Then $\operatorname{tg} \angle A B M$ $=$ . $\qquad$
Solution 1: As shown in Figure 1, let $AB=1, AM=x$, then $\operatorname{tg} \angle ABM=x$. The problem is converted into finding an equation about $x$. From $MD=1-x$, $DN=\frac{1}{2}$, we get
$$
MN=\sqrt{(1-x)^{2}+\frac{1}{4}}.
$$

Draw $NE \parallel BC$, intersecting $MB$ at $F$ and $AB$ at $E$. Since $N$ is the midpoint of $CD$, we have $EF=\frac{1}{2} AM=\frac{x}{2}$. Thus, $NF=1-\frac{x}{2}$.
But by the properties of parallel lines, we know
$$
\angle NFM=\angle CBM=\angle NMF.
$$

Thus, $NF=MN$.
We get $1-\frac{x}{2}=\sqrt{(1-x)^{2}+\frac{1}{4}}$.
Squaring and rearranging gives $(3x-1)(x-1)=0$.
But $M$ is different from $D$, so $x-1<0$. Thus, $x=\frac{1}{3}$, i.e., $\angle ABM=\frac{1}{3}$.

Analysis: The auxiliary line $NE \parallel BC$ can also be replaced by $NG \parallel MB$ intersecting $BC$ at $G$ (Figure 2) or $NH \parallel BM$ intersecting $AD$ at $H$ (Figure 3), leading to $BG=MN$, $MH=MN$, which has the same effect as (1).

Solution 2: As shown in Figure 4, rotate $\triangle ABM$ around $B$ by $90^{\circ}$ to get the right-angled $\triangle CBP$. Then $\angle BPC=\angle BMA=\angle CBM=\angle NMB$. Connect $MP$, by the equal base angles of the isosceles right-angled $\triangle BMP$, we get
$\angle NMP=\angle NPM$. Thus, $MN=NP=NC+CP=NC+AM$, i.e., $\sqrt{(1-x)^{2}+\frac{1}{4}}=\frac{1}{2}+x$.
We get $12x=4$. Solving gives $x=\frac{1}{3}$.
Analysis: This solution can eliminate the quadratic term in equation (2), solving a linear equation. If Figure 1 and Figure 4 are combined, a linear equation can be directly listed.
By $NF=NP$, we have
$1-\frac{x}{2}=\frac{1}{2}+x$.

Thus, $x=\frac{1}{3}$.
Solution 3: If one is limited to junior high school knowledge, using the coordinate method is also an option. As shown in Figure 5, establish a Cartesian coordinate system, set $BC$
$$
\begin{array}{l}
=BA=1, \\
AM=t, \\
\angle MBC=\alpha.
\end{array}
$$

Then
$M(t, 1), N\left(1, \frac{1}{2}\right)$,
$\operatorname{tg} \alpha=\frac{1}{t}$,
$\operatorname{tg} 2\alpha=\frac{M_y-N_y}{M_x-N_x}=\frac{1}{2(t-1)}$.

But $\operatorname{tg} 2\alpha=-\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^2 \alpha}$,
Thus, $\frac{1}{2(t-1)}=\frac{\frac{2}{t}}{1-\frac{1}{t^2}}$,
i.e., $4t(t-1)=t^2-1$.