Trigonometry/Cyclic Quadrilaterals and Ptolemy's Theorem
A cyclic quadrilateral is one where all four vertices lie on the same circle.
A quadrilateral is cyclic if and only if the two pairs of opposite angles each sum to 180º.
Draw the radii from two opposite vertices to the centre; they make two angles that must add to 360º. Since the angle at the circumference is half the angle at the centre, the sum of the opposite angles must be 180º. To prove the converse, consider the circumcircle of any three vertices; if the fourth vertex does not lie on this circle, opposite angles will not add to 180º.
With the above notation, if AB means the length of the line AB, etc., then
- AB.CD + BC.DA = AC.BD
Let AC and BD be drawn. Find a point X on AC such that ∠ABX = ∠CBD. Since ∠ABX + ∠CBX = ∠ABC = ∠CBD + ∠ABD, it follows that ∠CBX = ∠ABD.
△ABX is similar to △DBC, and △ABD is similar to △XBC. Thus AX/AB = CD/BD, and CX/BC = DA/BD, so
- AX·BD = AB·CD, and CX·BD = BC·DA
Adding these two equalities,
- AX·BD + CX·BD = AB·CD + BC·DA
- (AX+CX)·BD = AB·CD + BC·DA;
Since AX+CX = AC, the result follows.
Suppose ABCD is cyclic and is such that ABC is an equilateral triangle. Then regardless of where D is on the arc AC, DB = DA + DC.
A maltitude of a quadrilateral is a perpendicular to a side passing through the midpoint of the opposite side.
For a cyclic quadrilateral, the four maltitudes are concurrent. If this point is X and the centre of the circumcircle of the quadrilateral is O, then the centroid of the vertices bisects OX.
If the vertices in clockwise order are A, B, C and D, this means that the triangles ABC, BCD, CDA and DAB all have the same circumcircle and hence the same circumradius. The incentres of these four triangles always lie on the four vertices of a rectangle; these four points plus the twelve excentres form a rectangular 4x4 grid. If P, Q, R and S are the centres of arcs AB, BC, CD and DA respectively on the circumcircle, then the lines PR and QS are the perpendicular bisectors of the sides of the rectangle.
The four orthocentres form a quadrilateral congruent to the original; the centroids and the centres of the nine-point circles each form a quadrilateral similar to the original.