A cyclic quadrangle ABCD is called harmonic when the fourth vertex D is the 4-th harmonic of the three other A, B, C i.e. considering the points as complex numbers the cross-ratio (ABCD)=-1 (look at Complex_Cross_Ratio.html ).
[1] This is a characteristic property of harmonic quadrangles: The pole of each diagonal {U = (AB), V = (CD)} is contained in the other diagonal {CD, AB} respectively.
[2] For such a quadrangle, D can be determined as the intersection of the circumcircle and the line joining C and the pole (AB) of line AB with respect to the circumcircle.
[3] For such a quadrangle project the circumcenter O on AB at point M. Then triangles AMD, CBD and CMA are similar. In fact, the similarity of two out of the three triangles implies the similarity to the third and this fact is equivalent to the quadrangle being harmonic.
[4] Two opposite vertices, {A,B} say, are inverse with respect to the circle kCD passing through the other vertices {C, D} and orthogonal to the circumcircle. This is again a characteristic property of the harmonic quadrilateral.
[5] For harmonic quadrangles the circle kCD passing from {C,D} and orthogonally to the circumcircle is an Apollonius circle for the triangles ABC and ABD.
[6] A harmonic quadrangle is characterized by the fact that the products of lengths of opposite sides are equal.

Consider the polar (n) of the intersection point of the diagonals {AB, CD}. From general properties of cyclic quadrilaterals we know that (i) the intersection points of opposite sides {E, F} and the poles of the diagonals are four points on (n). To show that one diagonal, CD say, passes through the pole U of the other diagonal use the harmonicity. In fact, the harmonicity property of the quadrangle is equivalent to the fact that the tangent tA at A and the three lines {AD, AC, AN} make a harmonic bundle (AD, AC, AN, tA) = -1. This implies that the intersection poin U' of ta and DC is harmonic conjugate to N with respect to {D,C}. This identifies U' with U and proves the first part of [1]. The argument can be reversed to prove also the inverse. [2] is a cosequence of [1].
Let now M be the intersection point of the diagonal AB with UO. Since angle UMN is orthogonal and {C,D} are harmonic with respect to {U,N}, line MN is bisector of angle DMC. Taking the symmetrics D* of D and C* of C with respect to line (bisector) UM, we see that angle(ADM) = angle(CAB), since the arcs CB and AC* are equal. The statements in [3] follow from these remarks.
[4] follows from [1]. In fact, if the quadrangle is harmonic then circle (V, VD) is orthogonal to the circumcircle and {A,B} are inverse to this circle. Inversely, if this happens, then CD is the polar of V and use of [1] completes the argument.
[5] is equivalent to [4] and [6] equivalent to [5] since in that case, {C, D} being on the Apollonian circle will satisfy AC/CB = AD/DB.

Remark Notice that AB is a symmedian line of triangle ACD. This follows from the similarity of triangles ACM, ADM, which implies that the distances of M from the sides AC, AD are analogous to the lengths of these sides. A characteristic property of the symmedians. Hence the diagonals of the harmonic quadrangle coincide with some symmedians of its "partial" triangles ACD, ADB etc. (leaving out one vertex).

## 2. Harmonic quadrangle and squares

Let F be a Moebius transformation and A0B0C0D0 a square. Then applying the transformation to the vertices of the square we obtain a harmonic quadrangle or a harmonic division of four points on a line i.e. {A=F(A0), B=F(B0), C=F(C0), D=F(D0)} are the vertices of a harmonic quadrangle if the circumcircle of the square maps to a circle.

It is trivial to see that the vertices of a square satisfy the definition of harmonicity. Since Moebius transformations preserve the cross-ratio and map circles to circles (or lines) the image will be a quadrangle if the circumcircle of the square maps to a circle. Otherwise the four image-points will lie on a line and build a harmonic division.

Corollary Take a point P and draw the lines joining it with the vertices of a square. Then the second intersection points of these lines with the circumcircle of the square make a harmonic quadrangle.

The proof results by applying an inversion G (or anti-inversion, if P is inner to the circumcircle of the square) which gives the vertices {A, B, C, D} as images of the vertices of the square. The inversion is the one with respect to the circle orthogonal to the circumcircle of the square and centered at P. Since every inversion is the composition of a Moebius map and a reflexion the proof follows from the previous property.