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