As shown in Figure 2, in the square $A B C D$, $P$ is a point on the $\frac{1}{4}$ circular arc with $C$ as the center and $C B$ as the radius, and $P B > P D$. Point $Q$ is on the line segment $P B$, such that $P Q = P D$, and $A Q$ intersects $D P$ at point $R$. Find the value of $\frac{A R}{R Q}$.
Solve as shown in Figure 3, draw a perpendicular from point $B$ to $DR$, with the foot of the perpendicular at $H$. Draw a line through point $R$ parallel to $PB$ intersecting $HB$ at point $S$, and connect $HA$ and $BD$.

Since quadrilateral $ABCD$ is a square,
$\Rightarrow \angle ABC=\angle ADC=90^{\circ}$
$\Rightarrow AD, AB$ are tangents to $\odot C$.
Since $\angle BHD=\angle BAD=90^{\circ}$, therefore, points $B, H, A, D$ are concyclic.
Thus, $\angle ABH=\angle ADH=\angle PBD$.
Therefore, $\angle PBH=\angle DBA=45^{\circ}$, which means
$\angle HPB=45^{\circ}=\angle ABD=\angle AHD$
$\Rightarrow AH \parallel BP$.
Also, $\angle HAB=\angle PDB$, so
$\triangle HAB \backsim \triangle PDB$.
Thus, $\frac{AR}{RQ}=\frac{AH}{PQ}=\frac{AH}{PD}=\frac{AB}{BD}=\frac{\sqrt{2}}{2}$.