## 1. Circles on sides of parallelogram

Consider a parallelogram ABCD and draw two pairs of circles, each pair being symmetric with respect to the center O of the parallelogram (intersection of diagonals) and having chords two opposite sides. The other intersection points of these circles build a parallelogram A'B'C'D', which is also symmetric with respect to O.
[1] The angles of A'B'C'D' are the same with those of ABCD.
[2] The circumcircles of (A'BC) and (ABC') pass also through D', which is the intersection of circumcircles of (A'AD) and (C'CD).

The proof is an easy angle-chasing argument indicated in the figure above. First show that the angles at C' and D are equal. Then use part-1 and the symmetry of the figure to show that D'CBA is cyclic.

## 2. Four intersecting Euler circles

Consider the four triangles {ABC,DAB,DBC,DCA} defined by a basic triangle ABC and an additional point D.
The Euler circles of these four triangles intersect at a point P.

The proof follows immediately from section-1 by considering the middles {E,F,G,H,I,J} of the sides of the four triangles. The resulting configuration is the one of the preceding paragraph and its results apply to show the claim.

Remark If the four points are orthocentric i.e. {vertices of some triangle + its orthocenter}, then infinite many hyperbolas pass through these points. If the four points are non-orthocentric then there is just one rectangular hyperbola passing through these points and P is its center. This subject is discussed in OrthoRectangular.html and MiquelDual.html .