Re: [EMHL] Miquel Puzzle [SPOILER]
- On 1-06-02, Antreas P. Hatzipolakis <xpolakis@...> wrote:
>Let 1,2,3,4 be four lines forming theThe solution is based in two lemmata (well-known theorems):
>complete quadrilateral (1234).
>The Miquel point of (1234) is the common point
>of the circles (123), (234), (341), (412)
>[and also the circle (O(123)O(234)O(341)O(412))
>ie the circle passing through the four O's of the
>triangles (123), (234), (341), (412)]
>Now, define the Miquel point of (1234)
>by using only LINES; not circles.
Lemma #1 :
The orthocenters H(123), H(234), H(341), H(412)
of the four triangles (123), (234), (341), (412)
are collinear (Steiner Line of (1234)).
Let P be a point. The reflections of the line
HP in the sidelines of triangle ABC concur in
a point on the circumcircle of ABC.
The Miquel Point of the complete quadrilateral
(1234) is the point of concurrence of the reflections
of the Steiner line of (1234) in its sidelines 1,2,3,4.