(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 .

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