Cyclic Polygons

Cyclicity

A circumcyclic polygon has an in-circle touching all its sides. An incyclic polygon has a circle through all its vertices. Either of these polygons can be called cyclic polygons. If either circle exists it is uniquely defined. A polygon with incircle and circumcircle is bicyclic. All triangles are bicyclic. All regular polygons are bicyclic. An acyclic polygon has neither incircle nor circumcircle.

To find the circumcentre of an incyclic polygon: draw perpendicular bisectors (mediators) of its sides; these all meet at the circumcentre. To find the incentre of a circumcyclic polygon draw bisectors of the internal angles; these all meet at the incentre.

Given two circles, one inside the other, from any point on the outer circle we can draw a tangent to the inner circle which will meet the outer circle again. From this new point on the outer circle we can draw a further tangent to the inner circle, meeting the outer circle at a third new point. We can continue this process, thus generating a cyclic open polygon. If we continue the process long enough the polygon will close (or will do so to within a degree of accuracy).

Cyclic Quadrilaterals

A quadrilateral with angles ABCD and sides abcd, named in cyclic order, has a circumcircle if and only if A+C = B+D and has an incircle if and only if a+c = b+d. The first of these conditions (since A+B+C+D = 360°) implies A+C = B+D = 180°. [Euclid III.22]

The lines joining the midpoints of opposite arcs are at right angles. These lines are parallel to the bisectors of the angles formed by the diagonals. [Do diagonals of circumscribed quadrilateral cross at same point as the diagonals of inscribed quadrilateral? and are they at right angles?]

The product of the diagonals equals the sum of the products of the pairs of opposite sides: that is AC.BD = AB.DC + AD.BC. (Proof: Take Y on BD so that angle BAY = angle CAD. Then triangles BAC and YAD are similar and AB.CD = AC.BY. Adding these and since BY+YD = BD we get the result.???) [Ptolemy's theorem]

If ABCD are four points on a circle and AC and BD meet at Q, then AQ.QC = BQ.QD. Conversely if this is true the four points lie on a circle. QA.QC is equal to QL² where L is one end of the chord through Q perpendicular to OQ. If ABCD are on a circle and AD and BC meet at P outside the circle, then AP.DP = BP.CP. Conversely if this is true the four points lie on a circle. The value PA.PD is equal to PT² where PT is one of the tangents from P. [Euclid III 35-37]