Anti-Steiner Point. New proof?

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• Dear Geometers, Given a line w passing through the orthocenter H of a triangle ABC. We denote by a´, b´, c´the reflections of w in the sidelines BC, CA, AB,
Message 1 of 2 , Jan 29, 2004
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Dear Geometers,

Given a line w passing through the orthocenter H of a triangle ABC.
We denote by a´, b´, c´the reflections of w in the sidelines BC, CA,
AB, respectively. Then, the lines a´, b´, c´meet at one point, and
this point lies on the circumcircle of triangle ABC.

Solution:
Let J a point on the circumcircle of the triangle ABC and the points
P, Q, R the feet of the perpendiculars from J to the sides AB, BC, AC
of the triangle ABC. The points P, Q and R are collinear (line t),
(see, Advanced Euclidean Geometry, Roger A. Jonhson, paragraph 192,
page 137). P is the center of the circle whose diameter is the
segment JX. Q is the center of the circle whose diameter is the
segment JZ. R is the center of the circle whose diameter is the
segment JY. The centers of the three circles are collinear (line t)
and the circles passes through point J. The three circles form a
coaxal system of second type or elliptical (see Advanced Euclidean
Geometry Roger A. Jonhson, paragraph 54 page 36) whose line of
centers is the line t and the basic points are J and V. The radical
axis of the coaxal system is perpendicular to line t at point U. The
Simson line t of the point J bisects, at point D, the line joining
the point J to the orthocenter H (see Advanced Euclidean Geometry
Roger A. Jonhson paragraph 327 page 207). The circle of center D and
radius DH belong to coaxal system. We have
JX/JP=JV/JU=JH/JD=JZ/JQ=JY/JR=2. Thus the points X, V,H, Z and Y are
collinear (line w). The central similarity (homothety) with center J
and ratio 2 carries the line of centers of coaxal system (line t)
onto line w (see I. M. Yaglom, Geometric Transformation II, page 10).
The line t is parallel to line w. Under this transformation, the
points X, Z and Y are the images of the points P, Q and R,
respectively. Then, all lines passing through the orthocenter H of a
triangle is parallel to a Simson line of a point on the circumcircle.
The reflections of the line w in the sidelines AB, CA, BC carries the
line w onto lines a´, b´, c´, but we have JP=PX; JQ=QZ; JR=RY. When
the line w is reflected in sidelines AB, CA, BC the images X, Z and Y
of the points P, Q and R coincide with the point J. Then the anti-
Steiner point with respect to a triangle is the result of the
reflections of the images (that are in line w) of the feet of a
Simson line in sidelines of triangle.
Is this a new proof?
• Dear Deoclecio, ... Yes, of course, it is. But still, showing that every line through H is parallel to some Simson line, and that the Simson line of J bisects
Message 2 of 2 , Jan 29, 2004
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Dear Deoclecio,

In Hyacinthos message #9151, you wrote:

>> Given a line w passing through the orthocenter H of
>> a triangle ABC. We denote by a´, b´, c´the
>> reflections of w in the sidelines BC, CA, AB,
>> respectively. Then, the lines a´, b´, c´meet at one
>> point, and this point lies on the circumcircle of
>> triangle ABC.
>>
>> Solution:
>> Let J a point on the circumcircle of the triangle
>> ABC and the points P, Q, R the feet of the
>> perpendiculars from J to the sides AB, BC, AC of
>> the triangle ABC. The points P, Q and R are
>> collinear (line t), (see, Advanced Euclidean
>> Geometry, Roger A. Jonhson, paragraph 192, page
>> 137). P is the center of the circle whose diameter
>> is the segment JX. Q is the center of the circle
>> whose diameter is the segment JZ. R is the center
>> of the circle whose diameter is the segment JY.
>> The centers of the three circles are collinear
>> (line t) and the circles passes through point J.
>> The three circles form a coaxal system of second
>> type or elliptical (see Advanced Euclidean
>> Geometry Roger A. Jonhson, paragraph 54 page 36)
>> whose line of centers is the line t and the basic
>> points are J and V. The radical axis of the coaxal
>> system is perpendicular to line t at point U.
>> The Simson line t of the point J bisects, at point
>> D, the line joining the point J to the orthocenter
>> H (see Advanced Euclidean Geometry Roger A.
>> Jonhson paragraph 327 page 207). The circle of
>> center D and radius DH belong to coaxal system. We
>> have JX/JP=JV/JU=JH/JD=JZ/JQ=JY/JR=2. Thus the
>> points X, V,H, Z and Y are collinear (line w). The
>> central similarity (homothety) with center J and
>> ratio 2 carries the line of centers of coaxal
>> system (line t) onto line w (see I. M. Yaglom,
>> Geometric Transformation II, page 10). The line t
>> is parallel to line w. Under this transformation,
>> the points X, Z and Y are the images of the points
>> P, Q and R, respectively. Then, all lines passing
>> through the orthocenter H of a triangle is
>> parallel to a Simson line of a point on the
>> circumcircle.
>> The reflections of the line w in the sidelines AB,
>> CA, BC carries the line w onto lines a´, b´, c´,
>> but we have JP=PX; JQ=QZ; JR=RY. When the line w
>> is reflected in sidelines AB, CA, BC the images
>> X, Z and Y of the points P, Q and R coincide with
>> the point J. Then the anti-Steiner point with
>> respect to a triangle is the result of the
>> reflections of the images (that are in line w) of
>> the feet of a Simson line in sidelines of triangle.
>> Is this a new proof?

Yes, of course, it is. But still, showing that every
line through H is parallel to some Simson line, and
that the Simson line of J bisects the segment JH is
some work.

Note that the line through X, Y, Z and H is called
STEINER LINE of the point J with respect to triangle
ABC.

Sincerely,
Darij Grinberg
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