1) All the bicirculars having the same in and cir-cumcircle b, a have their diagonals pass through the same point E, which is the pole of the line e, which is the common radical axis of the (coaxal) circle-bundle I(a,b) generated by the two circles.

2) The locus of their centroids (middles of a segment joining two opposite sides) is a circle with diameter end-points the incenter and the middle of EF, F being the circumcenter.

3) The locus of the middles of their sides is an epitrochoid touching the incircle. The locus of the middles of the segments joining the middles of two adjacents sides is another epitrochoid.

In CyclicProjective.html has been proved that the circle with diameter EF, which is the locus of the middles of chordes of (a) passing through E, is orthogonal to the circle (g) with diameter OM. There has been noticed also that the line XY joining the middles of the diagonals of the quadrangle ABCD passes though N, center of (g). Thus X, Y are inverse with respect to (g). Extend XN to cut (g) at Z. Then (Z,L,Y,X) = -1 is a harmonic tetrad. Assume L is the other intersection point of XY with (g). Then, since OZ, OL are orthogonal they are the bisectors of angle YOX. But, from similarity of triangles OXC and OYD we see that OZ, OL are also bisectors of angle BOA. Thus L coincides with the intersection of the bisector of BOA with the circle (g). For bicentrics this is the incenter. Thus, line XY intersects EF at the incenter L. The argument proves that the centroid of the bicentric ABCD lies on a circle with diameter LJ, J being the middle of EF.

Document Newton2.html contains another proof of the fact that the Newton line of a circumscriptible passes through the incenter.

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