Example 3 As shown in Figure 4, given that the side length of square $A B C D$ is $1, P$ and $Q$ are two points inside it, and $\angle P A Q = \angle P C Q = 45^{\circ}$. Find the value of $S_{\triangle P A B} + S_{\triangle P C Q} + S_{\triangle Q A D}$. ${ }^{[3]}$
As shown in Figure 4, rotate $\triangle A Q D$ 90° clockwise around point $A$ to $\triangle A Q^{\prime} B$, and rotate $\triangle C Q D$ 90° counterclockwise around point $C$ to $\triangle C Q^{\prime \prime} B$. Connect $P Q^{\prime}$ and $P Q^{\prime \prime}$. Then,
$$
\begin{array}{l}
\triangle A P Q^{\prime} \cong \triangle A P Q, \triangle C P Q^{\prime \prime} \cong \triangle C P Q, \\
\angle A B Q^{\prime}=\angle A D Q, \angle C B Q^{\prime \prime}=\angle C D Q .
\end{array}
$$

Thus, $S_{\text {square } A B C D}=S_{\text {pentagon } A Q^{\prime} B C Q^{\prime \prime}}$.
Since $\angle A D Q + \angle C D Q = 90^{\circ}$, then
$$
\angle A B Q^{\prime} + \angle C B Q^{\prime \prime} = 90^{\circ}.
$$

Given $\angle A B C = 90^{\circ}$, we have
$$
\angle Q^{\prime} B Q^{\prime \prime} = 90^{\circ} + 90^{\circ} = 180^{\circ}.
$$

Therefore, points $Q^{\prime}$, $B$, and $Q^{\prime \prime}$ are collinear, and
$$
\begin{array}{l}
Q^{\prime} B = D Q = B Q^{\prime \prime}. \\
\text{Hence } S_{\triangle B P Q^{\prime}} = S_{\triangle B P Q^{\prime}} \\
\Rightarrow S_{\triangle P A B} + S_{\triangle Q A D} = S_{\text {quadrilateral } A Q^{\prime} B P} \\
\quad = \frac{1}{2} S_{\text {quadrilateral } A P Q^{\prime}} + \frac{1}{2} S_{\triangle P Q^{\prime} Q^{\prime}} \\
\Rightarrow S_{\triangle P A B} + S_{\triangle P C Q} + S_{\triangle Q A D} = \frac{1}{2} S_{\text {pentagon } Q^{\prime} Q^{\prime} Q C Q} \\
= \frac{1}{2} S_{\text {square } A B C D} = \frac{1}{2}.
\end{array}
$$