Given $ABCD$ is a square, $P$ is a point on side $BC$, and line $DP$ intersects the extension of $AB$ at point $Q$. If $DP^2 - BP^2 = BP \cdot BQ$, find the degree measure of $\angle CDP$ (do not use inverse trigonometric functions in your answer).

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Solution: As shown in Figure 2, let $\angle C D P=\angle P Q B=\theta$. Take a point $T$ on side $A B$ such that $B T=B P$, and connect $D T$ and $D B$. By symmetry, it is easy to see that
$$
D T=D P, \angle B D T=\angle B D P, \angle A D T=\angle C D P=\theta.
$$

From the given conditions, we have
$$
\begin{array}{l}
D T^{2}=D P^{2}=B P^{2}+B P \cdot B Q \\
=B T^{2}+B T \cdot B Q=B T \cdot(B T+B Q)=B T \cdot T Q,
\end{array}
$$

which means $\frac{D T}{B T}=\frac{T Q}{D T}$.
Since $\angle D T B=\angle Q T D$, we have
$\triangle D T B \backsim \triangle Q T D$.
Therefore, $\angle B D P=\angle B D T=\angle D Q T=\theta$.
Thus, $4 \theta=\angle A D C=90^{\circ}$, which means $\angle C D P=\theta=22.5^{\circ}$. Note: This problem can also be solved using the area method, trigonometric method, or algebraic method.