Problem:
A quadrilateral is inscribed in a circle of radius 2002​. Three of the sides of this quadrilateral have length 200. What is the length of its fourth side?
Answer Choices:
A. 200
B. 2002​
C. 2003​
D. 3002​
E. 500
Solution:
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 AB,BC, and CD are shorter than the radius, each of ∠AOB,∠BOC, and ∠COD is less than 60∘, so O is outside the quadrilateral ABCD. Let G and H be the intersections of AD with OB and OC, respectively. Because AD and BC are parallel, and △OAB and △OBC are congruent and isosceles, it follows that ∠ABO=∠OBC=∠OGH=∠AGB. Thus △ABG,△OGH, and △OBC are similar and isosceles with BGAB​=GHOG​=BCOB​=2002002​​=2​. Then AG=AB=200,BG=2​AB​=2​200​=1002​, and GH=2​OG​=2​BO−BG​=2​2002​−1002​​=100. Therefore AD=AG+GH+HD=200+100+200=(E)500​.
