[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.

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.

EulerCircleProperty.html

HyperbolaRectangular.html

MiquelDual.html

OrthoRectangular.html

RectangularAsProp.html

RectHyperbola.html

RectHypeCircumscribed.html

RectHypeThroughFourPts.html

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