Alas, I'm not familiar with how to get an overbar in this. So I will denote the conjugate of a complex number z by z'
Note that z and w are orthogonal iff z'*w is imaginary.
That is iff z'w = -zw'

Let the vertices of the quadrilateral be a, b, c and d in your diagram. Let the circumcircle of the quadrilateral be the unit circle. So |a| = 1 and so on.

Let m = (a + b + c + d)/2

Claim 1: m is the anticentre of abcd
Proof:
cc' = 1 = dd'
=> cc'-cd'+dc'-dd'=-c'c+c'd-d'c+d'd
=> (c+d)(c'-d') = -(c'+d')(c-d)
=>(m-(a+b)/2)(c'-d') = -(m'-(a'+b')/2)(c-d)
=> m-(a+b)/2 is orthogonal to c-d
=>m lies on a line through the midpoint of ab, perpendicular to cd.
The result follows by symmetry

Properties of the anticentre:
Claim 2: m lies on the nine-point circles of abc, bcd, cda and dab
Proof:
Let g be the centroid of abc. (g=(a+b+c)/3)
-d lies on the circumcircle of abc
The mapping f: z->(3g-z)/2 maps the circumcircle to the nine point circle of abc. But f(-d) = (a+b+c+d)/2 = m
So m lies on the nine point circle of abc
The rest follows once more by symmetry

Claim 3: The nine-point centres of abc, bcd, cda and dab are cyclic, with centre m
Proof:
This follows swiftly from the last claim, since each nine point circle has radius 1/2 and passes through m. So each nine point centre is a distance 1/2 from m

Claim 4: The quadrilateral formed by the orthocentres of abc, bcd, cda and dab is a rotation of abcd by pi about the anticentre of abcd
Proof: This rotation is the mapping g: z-> 2m-z
g(a) = a+b+c+d-a = b+c+d
Which is the orthocentre of bcd
the result follows, as before, by symmetry

that's all I can think of, but I'm sure there are many more.

Thankyou

sfwc
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