1) Its Euler circle passing through the center O of the hyperbola.

2) Its Orthocenter H lying also on the hyperbola.

3) The fourth intersection point N of the circumcircle with the hyperbola symmetric to H w.r. to O.

The clue fact is that the angle of two lines is equal to the angle of their conjugates (discussed in RectangularAsProp.html ). This implies that the angle of two chords is equal to the angle of the lines joining their middles with the center O of the hyperbola. Consider now triangle ABC inscribed in the rectangular hyperbola and draw its height AL. Define H to be the intersection point of AL with the hyperbola. We want to show that H is the orthocenter. Join H to the other vertices and define the middles G, I, E, D, M, F of the segments CH, CB, HB, ...etc.

It is well known that the quadrilateral DEFL is cyclic and its circumcircle is the Euler circle. By the clue-fact refered above, angle(DOF) = angle(DEF), hence O belongs to the Euler circle. This proves (1).

Consider now the middle M of AH and the intersection point M' of AH with the Euler circle. For M' we know that angle(EOM') = angle(ELM') = angle(EAM'). For M, by the clue-fact, we know that also angle(EOM) = angle(EAM') = angle(EAM). Thus M and M' make equal angles to OE, hence they coincide. The Euler circle passes through M and H is the orthocenter. This proves (2).

The third assertion is a consequence of the first two, since the Euler circle is homothetic w.r to the orthocenter by a factor of (1/2) and the hyperbola is symmetric w.r. to its center O.

By (3) of the first proposition, H is the orthocenter of the triangle. Since it is also the circumcenter, the triangle is equilateral.

AsymptoticTriangle.html

Hyperbola.html

HyperbolaAsymptoticProperty.html

HyperbolaAsymptotics.html

HyperbolaRectangular.html

Orthic.html

RectangularAsProp.html

RectHypeParaChords.html

RectHyperbola.html

RectHypeRelation.html

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