Construct any triangle ABC
![](assignment%208-1.png)
Construct
the Orthocenter H of triangle ABC
![](assignment%208-2.png)
Construct the nine point circles for each of the four resulting triangles
ABC, HBC, HAC, and HAB.
![](assignment%208-7.png)
I
conjecture that the nine-point 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)
![](9ptcircnumbers.png)
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 H--the
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 A--segment
C7 is perpendicular to AB, Point 5 is the foot of the altitude from
C--by 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 H--the
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 B--segment
C7 is perpendicular to AB, Point 2 is the foot of the altitude from
C--by 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 H--the
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 A--segment
B2 is perpendicular to AC, Point 5 is the foot of the altitude from
A--by construction A5 is perpendicular to BC.
Construct
the Orthocenters of each of the new triangles HBC, HAB, and HAC
![](assignment%208-4.png)
Conjecture:
The
orthocenter, H1, of triangle HAC appears to be the point B--a vertex of
the original triangle.
The
orthocenter, H2, of triangle HAB appears to be the point C--a vertex of
the original triangle.
The
orthocenter, H3, of triangle HBC appears to be the point A--a 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.
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