Q338 | Math Olympiad | Geometry | 2004 AMC 12A Problem 18 | Two Tangent Theorem

[@RealQinnMalloryu4]

(2)^2=4 360°ABCD/4 = 90°ABCD 3^30 3^5^6 3^5^13^2 1^1^13^2 3^2 (ABCD ➖ 3ABCD+2)

[@propofoldreams]

just for fun. let O be center of circle.
tan(OCB) = 1/2
thus tan(ECB) = 4/3
thus cos(DCE) = cos(90 - ECB) = sin(ECB) = 4/5
EC = 2 / cos(DCE) = 2 / (4/5) = 5/2