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?