Construct any triangle ABC
Construct
the Orthocenter H of triangle ABC
Construct the nine point circles for each of the four resulting triangles
ABC, HBC, HAC, and HAB.
I
conjecture that the ninepoint circle for each triangle is the same circle.
Proof:
The
nine point circle for a given triangle passes through the following nine
points:
 The midpoint
of each side of the triangle (3 points)
 The foot of each
altitude (3 points)
 The midpoints
of the segments from the vertex to the orthocenter (3 points)
The
above is constructed to be the nine point circle for the triangle ABC.
By construction the following follow:
 Point 1 is the
midpoint of segment AC
 Point 4 is the
midpoint of segment BC
 Point 8 is the
midpoint of segment AB
 Point 7 is the
foot of the altitude from C to segment AB
 Point 5 is the
foot of the altitude from A to segment BC
 Point 2 is the
foot of the altitude from B to segment AC
 Point 3 is the
midpoint of segment HC
 Point 6 is the
midpoint of segment HB
 Point 9 is the
midpoing of segment HA
Let us show this circle is
also the nine point circle for triangle HAC
 The circle passes through
the midpoint of each side of the triangle (Point 1 is the midpoint of
AC, Point 3 is the midpoint of HC, and Point 9 is the midpoint of HA)
 The point B is the orthocenter
of triangle HAC (see proof below); The circle passes through midpoint
of the segment from the orthocenter (B) to each of the vertices (Point
6 is the midpoint of HB; Point 4 is the midpoint of BC, Point 8 is the
midpoint of BA)
 The circle passes through
foot of each altitude. (Point 2 is the foot of the altitude from Hthe
segment from B to point is the altitude and H lies on this segment by
construction, Point 7 is the foot of the altitude from vertex Asegment
C7 is perpendicular to AB, Point 5 is the foot of the altitude from
Cby construction A5 is perpendicular to BC.)
Let us show this circle is
also the nine point circle for triangle HBC
 The circle passes through
the midpoint of each side of the triangle (Point 6 is the midpoint of
HB, Point 3 is the midpoint of HC, and Point 4 is the midpoint of BC)
 The point A is the orthocenter
of triangle HBC (see proof below); The circle passes through midpoint
of the segment from the orthocenter (B) to each of the vertices (Point
9 is the midpoint of HA; Point 1 is the midpoint of AC, Point 8 is the
midpoint of BA)
 The circle passes through
foot of each altitude. (Point 5 is the foot of the altitude from Hthe
segment from A to point 5 is the altitude and H lies on this segment
by construction, Point 7 is the foot of the altitude from vertex Bsegment
C7 is perpendicular to AB, Point 2 is the foot of the altitude from
Cby construction B2 is perpendicular to AC.)
Let
us show this is also the nine point circle for triangle HAB
 The circle passes through
the midpoint of each side of the triangle (Point 6 is the midpoint of
HB, Point 9 is the midpoint of HA, and Point 8 is the midpoint of BA)
 The point C is the orthocenter
of triangle HAB (see proof below); The circle passes through midpoint
of the segment from the orthocenter (B) to each of the vertices (Point
4 is the midpoint of CB; Point 3 is the midpoint of HC, Point 1 is the
midpoint of AC)
 The circle passes through
foot of each altitude. (Point 7 is the foot of the altitude from Hthe
segment from C to point 7 is the altitude and H lies on this segment
by construction, Point 2 is the foot of the altitude from vertex Asegment
B2 is perpendicular to AC, Point 5 is the foot of the altitude from
Aby construction A5 is perpendicular to BC.
Construct
the Orthocenters of each of the new triangles HBC, HAB, and HAC
Conjecture:
The
orthocenter, H1, of triangle HAC appears to be the point Ba vertex of
the original triangle.
The
orthocenter, H2, of triangle HAB appears to be the point Ca vertex of
the original triangle.
The
orthocenter, H3, of triangle HBC appears to be the point Aa vertex of
the original triangle.
Proof:
Show
B is the intersection of the altitudes of triangle HAC. By construction
the altidude from vertex H is on the line that passes through HB. By construction,
the line through HC is perpendicular to the line AB. So the altitude from
vertex A lies on the line that passes trhough AB.
By Construction, the line that passes through CB is perpendicular to the
line that passes through AH. So, the altitude from vertex C lies on the
line that passes through CB. B is on each of the lines containing the
altitudes, so B is the orthocenter.
Show
C is the intersection of the altitudes of triangle HAB. By construction
the altidude from vertex H is on the line that passes through HC. By construction,
the line through HA is perpendicular to the line BC. So the altitude from
vertex B lies on the line that passes trhough BC.
By Construction, the line that passes through BH is perpendicular to the
line that passes through AC. So, the altitude from vertex A lies on the
line that passes through AC. C is on each of the lines containing the
altitudes, so C is the orthocenter.
Show
A is the intersection of the altitudes of triangle HBC. By construction
the altidude from vertex H is on the line that passes through HA. By construction,
the line through HC is perpendicular to the line AB. So the altitude from
vertex B lies on the line that passes trhough AB.
By Construction, the line that passes through BH is perpendicular to the
line that passes through CA. So, the altitude from vertex C lies on the
line that passes through CA. A is on each of the lines containing the
altitudes, so A is the orthocenter.
