The orthocycle (c) of a quadrangle q = ABCD is the circle with diameter FG, where F, G are the intersection points of the opposite sides of q. The quadrilateral is cyclic if and only if the orthocycle is orthogonal to the circle f passing through the three points E, X, Y which are the diagonals-intersection-point and the two middles of the diagonals correspondingly.
(1) If q is cyclic, then by Newton's theorem X, Y and H, the middle of FG are collinear (see Newton.html ). From the basic construction of harmonic conjugate points, F and E are on the polar of G w.r. to (c). Besides the circle (f) through E, X, Y is the locus of middles of chords of (c) passing through E. From the two right angled triangles FPG and EPO follows that the circles (f) , (c) intersect orthogonally.
(2) Inversely, if the two circles intersect orthogonally, then Y, X are inverse w.r. to (c). Since the same is true with R, Q, there is a circle (a) passing through the four points X, Y, R and Q. Then we have ER*EX = EY*EQ (i). But from the general properties of the complete quadrilaterals we have also that (Q,E,C,A)=-1 is a harmonic division, hence EC*EA = EQ*EY (ii). Analogously (R,E,B,D)=-1 implies EB*ED = ER*EX (iii). Relations (i) to (iii) imply that EB*ED = EC*EA, showing that A, B, C, D are on a circle.
For the consequence of harmonicity (Q,E,C,A)=-1 = > EC*EA = EQ*EY see the general properties of harmonic tetrades handled in the file Harmonic.html .