- Jan 29, 2004Dear Deoclecio,

In Hyacinthos message #9151, you wrote:

>> Given a line w passing through the orthocenter H of

Yes, of course, it is. But still, showing that every

>> 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?

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 - << Previous post in topic