$20 \cdot 22$ In circle $O$, diameter $AB$ is 10 units, point $C$ is on $AB$ 4 units from $A$, point $D$ is on $AB$ 4 units from $B$, and $P$ is any point on the circumference of the circle. Then the broken path from $C$ through $P$ to $D$
(A) is the same length for all positions of $P$.
(B) is more than 10 units for all positions of $P$.
(C) cannot exceed 10 units.
(D) is shortest when $\triangle CPD$ is a right triangle.
(E) is longest when $PC = PD$.
[Solution] Connect $PO$, by the median theorem

Also,
$$
\begin{array}{l}
P C^{2}+P D^{2}=2\left(5^{2}+1^{2}\right)=52, \\
P C \cdot P D \leqslant \frac{1}{2}\left(P C^{2}+P D^{2}\right),
\end{array}
$$

Therefore, the maximum value of $P C \cdot P D$ is 26, and it is achieved when $P C=P D=\sqrt{26}$.
$$
\text { Also, }(P C+P D)^{2}=52+2 P C \cdot P D \leqslant 52+52=104 \text {, }
$$

Thus, $P C+P D \leqslant 2 \sqrt{26}$,
Hence, $P C+P D$ is longest when $P C=P D=\sqrt{26}$.
Therefore, the answer is $(E)$.