Complete Quadrilateral
Complete quadrilateral
Let four lines made four triangles of a complete quadrilateral. In the diagram these are
One can see some of the properties of this configuration and their proof using the following links.
Radical axis
Let four lines made four triangles of a complete quadrilateral. In the diagram these are
Let points and
be the orthocenters of
and
respectively.
Let circles and
be the circles with diameters
and
respectively.
Prove that Steiner line
is the radical axis of
and
Proof
Let points and
be the foots of perpendiculars
and
respectively.
Denote power of point
with respect the circle
Therefore power of point
with respect these three circles is the same. These points lies on the common radical axis of
and
Steiner line
is the radical axis as desired.
vladimir.shelomovskii@gmail.com, vvsss
Newton–Gauss line
Let four lines made four triangles of a complete quadrilateral.
In the diagram these are
Let points and
be the midpoints of
and
respectively.
Let points and
be the orthocenters of
and
respectively.
Prove that Steiner line is perpendicular to Gauss line
Proof
Points and
are the centers of circles with diameters
and
respectively.
Steiner line is the radical axis of these circles.
Therefore as desired.
vladimir.shelomovskii@gmail.com, vvsss
Shatunov line
Let the complete quadrilateral ABCDEF be labeled as in the diagram.
Let points be the orthocenters and points
be the circumcenters of
and
respectively.
Let bisector cross bisector
at point
Let bisector
cross bisector
at point
Prove that
a) points and
lie on circumcircle of
b) line is symmetric to Steiner line with respect centroid of
I suppose that this line was found by a young mathematician Leonid Shatunov in November 2020. I would be grateful for information on whether this line was previously known.
Proof
a) Points and
lies on bisector of
points
and
lies on bisector of
circle
Similarly circle
as desired.
b) Let and
be midpoints of
and
respectively.
It is clear that is centroid of
(midline of trapezium
(midline of trapezium
is parallelogram.
Similarly one can prove that point the midpoint of
is symmetric to
with respect
Therefore line coincide with Steiner line and line
is symmetric to Steiner line with respect
and is parallel to this line.
vladimir.shelomovskii@gmail.com, vvsss