A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$
Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$

$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$

From there, $\triangle{OAB} \sim \triangle{ABE}$, thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$. Since $\frac{OA}{AB} = \frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$.
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}$