24. Given a cyclic quadrilateral $ABCD$ with the radius of the circumscribed circle being $200 \sqrt{2}, AB=BC=CD=200$. Then the length of side $AD$ is ( ).
(A) 200
(B) $200 \sqrt{2}$
(C) $200 \sqrt{3}$
(D) $300 \sqrt{2}$
(E) 500
24. E.

By the cosine rule, we have
$$
\begin{array}{l}
\cos \angle B O C=\frac{B O^{2}+O C^{2}-B C^{2}}{2 B O \cdot O C} \\
=\frac{(200 \sqrt{2})^{2}+(200 \sqrt{2})^{2}-200^{2}}{2 \times 200 \sqrt{2} \times 200 \sqrt{2}}=\frac{3}{4} .
\end{array}
$$

Given $A B=B C=C D$, we know $\angle B O C=\angle C O D$.
Then $\sin \angle C O D=\frac{\sqrt{7}}{4}$
$\Rightarrow \cos \angle B O D=2 \cos ^{2} \angle B O C-1=\frac{1}{8}$.
Thus, $B D^{2}=B O^{2}+O C^{2}-2 B O \cdot O C \cos \angle B O D$
$=200^{2} \times 2+200^{2} \times 2-2 \times 200 \sqrt{2} \times 200 \sqrt{2} \times \frac{1}{8}$
$$
=\frac{7}{2} \times 200^{2} \text {. }
$$

By Ptolemy's theorem, we have
$$
\begin{array}{l}
A B \cdot C D+B C \cdot A D=A C \cdot B D \\
\Rightarrow 200^{2}+200 A D=\frac{7}{2} \times 200^{2} \\
\Rightarrow A D=500 .
\end{array}
$$