Consider triangle t=ABC and a point D on its circumcircle. Project D on the sides of (t) and define the Simson-Wallace line W(D) (see Simson.html ) through these projections: E, F and G. Extend DE to define its intersection point H with the circumcircle. Then CH is parallel to the Simson line W(D).
The proof follows at once from the fact that quadrangles DFEB and DCHB are cyclic. Hence angle(CHD) = angle(CBD) = angle(HEF). Showing that EF and CH are parallel.
This has an interesting consequence studied in SimsonProperty2.html . Remark Another application is the determination of a point D, such that its Simson line S(D) has a given direction:
Draw from C a line CH in the given direction and determine point H. Then from H draw the orthogonal to the oposite side (AB) intersecting the circle at a second point D. This is the desired point.
From a point P on the circumcircle of triangle ABC draw PDF orthogonal to side BC, F being the second intersection point with the circumcircle. Extend the altitude HA by AG = DF, H being the orthocenter. Then GP = HD or equivalently HDPG is a parallelogram.
The proof follows from the equality of triangles DEF and GPA which is trivial and the fact that E is symmetric to H with respect to BC. Remark Since by the previous section DG is the Simson-Wallace line of P, the middle M of PH is on this line and simultaneously on the Euler circle of ABC, which is homothetic to the circumcircle with respect to H and in ratio 1/2. This gives another argument to prove the property of the Steiner line parallel to DG (see SteinerLine.html ).
This argument is used also to prove a property of orthopoles (see Orthopole.html ).
![[alogo]](alogo.png){kind=small}
1. A parallel to the Simson-Wallace line ![[0_0]](SimsonProperty_files/SimsonProperty_0_0_0.png)
![[0_0]](SimsonProperty_files/SimsonProperty_1_0_0.png)