Example 4 As shown in Figure 3, the radius of $\odot O$ is $R, C$ and $D$ are two points on the circumference of the circle on the same side of the diameter $A B$, the degree measure of $\overparen{A C}$ is $96^{\circ}$, and the degree measure of $\overparen{B D}$ is $36^{\circ}$. A moving point $P$ is on $A B$. Find the minimum value of $P C + P D$.
(1997, Shaanxi Province Junior High School Mathematics Competition)
Solution: Let $D^{\prime}$ be the point symmetric to point $D$ with respect to the diameter $A B$, and connect $C D^{\prime}$ intersecting $A B$ at $P$. By the symmetry of the circle, point $D^{\prime}$ lies on the circle. Therefore,
$$
P D+P C=P D^{\prime}+P C .
$$

Since $P D^{\prime}+P C$ is the shortest at this time,
$P D+P C$ is the shortest.
Connect $C O$ and $D^{\prime} O$, and draw $O M \perp C D^{\prime}$ at $M$.
Since $\overparen{A C}=96^{\circ}, \overparen{B D}=36^{\circ}$,
thus, $\overparen{C D}=48^{\circ}$.
Therefore, the degree of $\overparen{C D^{\prime}}$ is $48^{\circ}+36^{\circ} \times 2=120^{\circ}$.
Hence, $\angle C O D^{\prime}=120^{\circ}$,
$$
\angle C O M=\frac{1}{2} \angle C O D^{\prime}=60^{\circ} \text {. }
$$

In the right triangle $\triangle C O M$,
$$
\begin{array}{l}
C O=R, C M=C O \cdot \sin 60^{\circ}=\frac{\sqrt{3}}{2} R, \\
C D^{\prime}=2 C M=\sqrt{3} R .
\end{array}
$$

Therefore, the minimum value of $P C+P D$ is $\sqrt{3} R$.