7859Re: CONCURRENT EULER LINES (was: Thebault point)
- Sep 11, 2003Dear Antreas,
In Hyacinthos message #7858, you wrote (partly):
>> Let ABC be a triangle and AHa, BHb, CHc its altitudes.^^^^^^^^^^^^^^^^^^^
>> Special case #1:
>> Let Ab, Ac be the reflections of A in Hc, Hb, resp.
>> Let Bc, Ba be the reflections of B in Ha, Hc, resp.
>> Let Ca, Cb be the reflections of C in Hb, Ha, resp.
>> The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb
I guess you mean AAbAc, BBcBa, CCaCb.
The Euler lines of these triangles concur since the
orthocenter H of triangle ABC is their common
>> concur (at H of ABC).Yes. In fact, the nine-point circle of a triangle
>> Problem: Are their Nine Point Circles concurrent?
ABC is the reflection of the circumcircle of triangle
AB'C' in the line B'C', where A', B', C' are the
midpoints of the sides BC, CA, AB. This yields that
the nine-point circle of triangle AAbAc is the
reflection of the circumcircle of triangle AHbHc in
the line HbHc. Similarly for the other two nine-point
circles. Since the circles AHbHc, BHcHa, CHaHb concur
at H, the three nine-point circles concur at the
antigonal conjugate of H with respect to triangle
NOTE. Antigonal conjugates? If P is a point in the
plane of a triangle ABC, then the reflections of the
circles PBC, PCA, PAB in the sidelines BC, CA, AB
have a common point. This point is called antigonal
conjugate of P with respect to triangle ABC. But I
see (Hyacinthos message #7826) that you know this,
maybe with another name.
The point where your nine-point circles concur is
not in the ETC.
>> Generalization:Yes, they are: the proof is simple. Moreover, if
>> Let Pac, Pab be points on AB, AC resp. such that:
>> APac / AHc = APab / AHb = t
>> Similarly we define the points Pba, Pbc; Pcb, Pca
>> The Euler Lines of the Triangles APabPac, BPbcPba,
>> CPcaPcb are concurrent
J = X(125) is the intersection of the Euler lines of
triangles AHbHc, BHcHa, CHaHb, then the Euler lines
of triangles APabPac, BPbcPba, CPcaPcb meet at the
point W lying on the line HJ and dividing the segment
HJ in the ratio
-- = ---.
Remark that the line HJ passes through X(74).
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