Cyclic from non-cyclic (Posted on 2008-07-18)

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NOTATION: Let a = |A'B'|, b = |B'C'|, c = |C'D'|, and d = |D'A'| Let P(r) denote the circle with center P and radius r. So as an example - A(|AB|) denotes a circle with center A and passing through point B. ANALYSIS: Clearly, if we can locate point C with respect to line segment AD, then point B will be one of the intersections of A(a) and the circumcircle of triangle CDA. We will assume ABCD to be cyclic in order to find the location of point C in relation to the line segment AD. Let point M be on the extention of side DA such that angle DCM equals angle BAC. Triangles CDM and ABC are similar since angle CDM equals angle ABC (Why?). Therefore, |DM| |BC| b*c ------ = ------ or |DM| = ----- |CD| |AB| a Clearly point M is constructible. Also, |AC| |AB| a ------ = ------ = --- (1) |CM| |CD| c Point C lies on circle D(c) and from equation (1) it also lies on the circle whose diameter is IE (where I and E are the interior and exterior centers of similitude of circles A(a) and M(c) - See here and here). Note: If a = c, then the "circle" becomes the perpendicular bisector of line segment AM. CONSTRUCTION: Construct point M on the extension of line segment AD such that |DM| = b*c/a. Construct A(a) and M(c). Construct a line through A, perpendicular to line AM, intersecting A(a) at points A1 and A2. Construct a line through M, perpendicular to line AM, intersecting M(c) at points M1 and M2. Labeling should be such that A1 and M1 are on the same side of line AM. The lines A1M2 and A1M1 intersect line AM at points I and E respectively. Construct the circle with diameter IE ( or if a = c, the perpendicular bisector of line segment AM) intersecting D(c) at two points. Pick one as point C. Construct the circumcircle of triangle CDA intersecting A(a) at two points. Pick B as the one that makes ABCD convex.