## 1. Miquel dual configuration

Under this name I understand the configuration resulting from a triangle by drawing three arbitrary lines, one through each vertex of the triangle.
There are some remarkable analogies between this and the usual Miquel configuration, resulting by taking three arbitrary points on the sides of the triangle. A discussion on this topic is contained in Miquel_Point.html . In the present configuration hold the following properties.
[1] Let {A',B',C'} be arbitrary points on line-sides respectively {BC,CA,AB} and {A1,B1,C1} the intersection points of the resulting lines {AA',BB',CC'}. Then circles (ABC1), (BCA1), (CAB1) pass all through a point O.
[2] Every other tripple of lines {AB2,BC2,CA2} resulting by taking {A2,B2,C2} respectively on the circles {(BCA1), (CAB1),(ABC1)}, so that these lines are equal inclined respectively to {AA',BB',CC'} define in analogous way the same point O. Inversely, every other three lines defining in the same way the same point O, are equal inclined respectively to lines {AA',BB',CC'}.
[3] Triangles A1B1C1 and A2B2C2 are always similar to each other and similar also to triangle A0B0C0, formed by the centers of circumcircles of triangles {(BCA1), (CAB1), (ABC1)}.

The clue for all these properties lies in the following two relations of angles.
First, considering the intersection O of circumcircles (BA1C) and (CB1A), angle(AOC) = angle(AB1C) = pi-angle(C0B0A0). Analogous relations hold for the other angles around O.
Second, angle(C'CA2) = angle(A'AB2) = angle(B'BC2).
Remark-1 The configuration is just another aspect of the well known pivoting of a triangle (like A2B2C2) around a fixed triangle, so that the pivoting triangle remains similar to itself. This is discussed in Similarity.html .
Remark-2 Special configurations deliver remarkable points O. For example, selecting {A',B',C'} on the sides, such that angles {B1AC,C1BA,A1CB} are all equal implies that triangles A2B2C2 is similar to ABC. The corresponding point O is the 2nd Brocard point of all these triangles including ABC.
Remark-3 The angles around O being equal to the exterior angles of triangle A1B1C1 determine its shape. Thus, if O takes the position of the orthocenter the corresponding A1B1C1 is similar to ABC. If O takes the position of the centroid then A1B1C1 becomes similar to the triangle of medians. If O takes the position of the circumcenter then A1B1C1 is similar to the orthic triangle etc..

## 2. Other analogies with Miquel pivot

There are some configurations bearing analogies to corresponding configurations involving the Miquel pivot. Among the more important is the case in which lines {AA',BB',CC'} pass through the same point D. The following figure shows this configuration.
[1] 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.
[2] The unique rectangular parabola passing through four non-orthocentroidal points {A,B,C,D} has its center at P.

The statement on the four Euler circles is proved in EulerCirclesFour.html . The other assertion follows from the general properties of triangles inscribed in rectangular hyperbolas (see RectHypeCircumscribed.html ). Any triangle inscribed in a rectangular hyperbola has its center on its Euler circle. Hence in our case P is the center of the hyperbola.