## 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]