AMC10 2016 A

Answer (E): Let $ABCD$ be the given quadrilateral inscribed in the circle centered at $O$, with $AB=BC=CD=200$, as shown in the figure. Because the chords $\overline{AB}$, $\overline{BC}$, and $\overline{CD}$ are shorter than the radius, each of $\angle AOB$, $\angle BOC$, and $\angle COD$ is less than $60^\circ$, so $O$ is outside the quadrilateral $ABCD$. Let $G$ and $H$ be the intersections of $\overline{AD}$ with $\overline{OB}$ and $\overline{OC}$, respectively. Because $\overline{AD}$ and $\overline{BC}$ are parallel, and $\triangle OAB$ and $\triangle OBC$ are congruent and isosceles, it follows that $\angle ABO=\angle OBC=\angle OGH=\angle AGB$. Thus $\triangle ABG$, $\triangle OGH$, and $\triangle OBC$ are similar and isosceles with $\dfrac{AB}{BG}=\dfrac{OG}{GH}=\dfrac{OB}{BC}=\dfrac{200\sqrt{2}}{200}=\sqrt{2}$. Then $AG=AB=200$, $BG=\dfrac{AB}{\sqrt{2}}=\dfrac{200}{\sqrt{2}}=100\sqrt{2}$, and $GH=\dfrac{OG}{\sqrt{2}}=\dfrac{BO-BG}{\sqrt{2}}=\dfrac{200\sqrt{2}-100\sqrt{2}}{\sqrt{2}}=100$. Therefore $AD=AG+GH+HD=200+100+200=500$.