For all possible relative positions of {A',B',C'} this follows by an easy angle-chasing argument as the one shown in the above figure.

(i) One point, C' say, coincides with one vertex, A say (first figure below).

(ii) The three points {A',B',C'} are on a line (second figure below).

The name

This because the three triangles {PA'A'',PB'B'',PC'C''} are similar. Consequently the triangles A'B'C', A''B''C'' are also similar. Thus, for varying fi, one can consider that triangle A'B'C' is pivoting around P, its vertices gliding on the side-lines of ABC.

In the above figure C' is considered coiniciding with A. The proof of the property is again an easy angle-chasing argument.

In fact, in this case the figure has two aspects :

(i) triangle ABC + intersecting line A'B'C'.

(ii) triangle A'BC' + intersecting line AB'C.

In the first case we have coincidence at P of the circles (AB'C'), (BC'A') and (CA'B').

In the second case we have coincidence at P of the circles (A'B'C), (BCA) and (C'AB').

A particular case of the last corollary is the one in which the three equal angles coincide with a right angle. Then line A'B'C' is a

Inversely, if points {C',A',B'} are collinear and set respectively on the sides {AB,BC,CA} of triangle ABC, then there is a point P such that {PC',PA',PB'} respectively are equal inclined to these sides and P is on the circumcircle of ABC.

The circumcircles of the four triangles formed by any three of the sides of a quadrilateral intersect at a point P. We call it the

To build such a triangle we leave out a side (line) of the quadrangle and build the triangle having sides the remaining lines. This figure has many properties and applications. Some of them are discussed below.

Leaving one of the O

By the conceptual symmetry of the configuration it suffices to show it for one triangle, O

First assertion follows directly from property-1. For example, leaving out O

The other assertion follows also easily by remarking, for example, that O

This follows from an easy angle chasing argument and property-1.

[2] The orthocenters of these four triangles are collinear and their support-line L' is parallel to L and d(P,L')=2d(P,L).

First assertion follows immediately from the fact that two sides of the quadrangle participate both in two triangles out of the four and from corollary-3 of section 2.

Second assertion is a consequence of the first and the general fact that a line L' parallel to a Simson-Wallace line L(P) at double distance from P passes through the orthocenter of the respective triangle (see SteinerLine.html ).

[2] The focus of the parabola is the Miquel point P of the quadrilateral formed by the sides of the quadrilateral. The tangent at the vertex of the parabola is the Simson-Wallace line of P with respect to any of the four triangles created by the sides of the quadrilateral.

[1] Is a consequence of the general property of conics to be uniquely defined through five lines in general position (i.e. no three of them passing through a point). The fifth missing line is the line at infinity to which every parabola is tangent.

[2] Is a consequence of the general property of triangles whose sides are tangents to a parabola. Their circumcircle passes through the focus and their Simson-Wallace line of the focus is the tangent at the vertex of the parabola (their orthocenter then lying on the directrix see ParabolaChords.html ).

Harmonic.html

Menelaus.html

MiquelDual.html

Newton.html

Newton2.html

OrthocyclicChar.html

ParabolaChords.html

SteinerLine.html

Tangent4Lines.html

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