Simson line
In geometry, given a triangle ABC and a point P on its circumcircle, the three closest points to P on lines AB, AC, and BC are collinear.
Proof
In the shown diagram, we draw additional lines and
. Then, we have cyclic quadrilaterals
,
, and
. (more will be added)
Simson line (main)
Let a triangle and a point
be given.
Let and
be the foots of the perpendiculars dropped from P to lines AB, AC, and BC, respectively.
Then points and
are collinear iff the point
lies on circumcircle of
Proof
Let the point be on the circumcircle of
is cyclic
is cyclic
is cyclic
and
are collinear as desired.
Proof
Let the points and
be collinear.
is cyclic
is cyclic
is cyclis as desired.
vladimir.shelomovskii@gmail.com, vvsss
Problem
Let the points and
be collinear and the point
Let and
be the circumcenters of triangles
and
Prove that lies on circumcircle of
Proof
Let and
be the midpoints of segments
and
respectively.
Then points and
are collinear
is Simson line of
lies on circumcircle of
as desired.
vladimir.shelomovskii@gmail.com, vvsss