This write up is for problems Assignment 8.

**Altitudes and Orthocenters**
by
**Brad Simmons**

To begin our exploration and discussion,
we construct the orthocenter (H) of triangle ABC. The **orthocenter**
of a triangle is the common intersection of the three lines containing
the altitudes. An **altitude **is a perpendicular segment from
a vertex to the opposite side of the triangle.

We can construct the orthocenter (
H') of triangle HBC. The image below shows H' and point A coincide.

We can construct the orthocenter (
H') of triangle HAB. The image below shows H' and point C coincide.

We can construct the orthocenter (
H') of triangle HAC. The image below shows H' and point B coincide.

For a GSP sketch that can be manipulated
please
click here.

**Given:**
point H is the orthocenter of triangle ABC
**Prove: **point
A is the orthocenter of triangle HBC
1. Segment AP is an altitude of triangle
ABC since line AP passes through the orthocenter H of triangle
ABC.
2. Segment AP is perpendicular to segment
BC because an altitude is a segment that joins a vertex with the
opposite side of a triangle and is perpendicular to that opposite
side.
3. Angle APB is a right angle because
segment AP is perpendicular to segment BC.
4. Point H lies on segment AP because
an altitude of a triangle passes through the triangle's orthocenter.
5. Angle HPB is a right angle because
segment HP is perpendicular to segment BC.
6. Segment HP is an altitude of triangle
HBC since it is a segment perpendicular to side BC and passing
through the vertex H.
7. Points A, H, and P are collinear
because the orthocenter H lies on the altitude AP.
8. Point A lies on the extension of
the altitude HP. (Altitude HP of triangle HBC).
9. Segment BG is an altitude of triangle
ABC since it passes through the orthocenter H of triangle ABC.
10. Segment BG is perpendicular to
segment AC because an altitude is a segment that joins a vertex
with the opposite side of a triangle and is perpendicular to that
opposite side.
11. Points B, H, G are collinear because
the orthocenter H lies on the altitude BG.
12. Angle HGC is a right angle because
segment AC is perpendicular to segment BG.
13. Segment CG is an altitude of triangle
HBC since it is a segment perpendicular to the extension of side
BH (at point G) and passing through the vertex C.
14. Point A lies on the extension of
the altitude CG since points A, C, and G are collinear.
15. By similar reasoning it can be
proved that Point A lies on the extension of segment BI which
is an altitude of triangle HBC.
16. Point A is the orthocenter of triangle
HBC since it is the common intersection of the lines containing
the three altitudes of triangle HBC.

By similar reasoning it can be proved
that Point B is the orthocenter of triangle HAC.

Furthermore, by similar reasoning it
can be proved that Point C is the orthocenter of triangle HAB.

Now if we construct the circumcircles
of triangles ABC, HBC, HAB, and HAC as shown below, we can explore
the behavior of the circumcircles if we manipulate the orthocenter
(H) or and of the vertices of triangle ABC. If we drag the point
H out to where the point C is now located, then point C will move
to where point H is now located and the red circumcircle and black
circumcircle will switch places as well. (Try this by clicking
on the GSP sketch link below this image.) Furthermore, if we move
point H out to the location of point A, then the black circumcircle
and blue circumcircle will switch places. What will happen if
we move point H out to the location of point B? Will the black
circumcircle and green circumcircle switch places?

For a GSP sketch that can be manipulated
please click
here.

The image below shows the nine point
circle (red) and orthocenter (H) of triangle ABC.

Now look at the nine point circle (red)
of triangle HBC.

Now look at the nine point circle (red)
of triangle HAC.

Is this nine point circle (red) also
the same for triangle HAB?

It appears from the images above that
all four triangles ABC, HBC, HAC, and HAB share the same nine
point circle. With the help of GSP we can construct the nine point
circle for all four triangles in the same sketch and then manipulate
the sketch to examine other acute, obtuse or right triangles.
For a GSP sketch that can be manipulated
please click
here.

Given: point
H is the orthocenter of triangle ABC.
Prove: triangle
ABC and triangle HBC share the same nine point circle.
1. For triangle ABC, point L is the
midpoint of the segment from vertex A to the orthocenter H. For
triangle HBC, point L is the midpoint of the segment from vertex
H to the orthocenter A.
2. For triangle ABC, point K is the
midpoint of the segment from vertex B to the orthocenter H. For
triangle HBC, point K is the midpoint of side HB.
3. For triangle ABC, point J is the
midpoint of the segment from vertex C to the orthocenter H. For
triangle HBC, point J is the midpoint of side HC.
4. For triangle ABC, point I is the
foot of the altitude CI. For triangle HBC, point I is the foot
of the altitude BI.
5. For triangle ABC, point P is the
foot of the altitude AP. For triangle HBC, point P is the foot
of the altitude HP.
6. For triangle ABC, point G is the
foot of the altitude BG. For triangle HBC, point G is the foot
of the altitude CG.
7. For triangle ABC, point D is the
midpoint of the side AC. For triangle HBC, point D is the midpoint
of the segment from vertex C to the orthocenter A.
8. For triangle ABC, point F is the
midpoint of the side AB. For triangle HBC, point F is the midpoint
of the segment from vertex B to the orthocenter A.
9. For triangle ABC, point E is the
midpoint of the side BC. For triangle HBC, point E is the midpoint
of the side BC.
10. The points L, K, J, I, P, G, D,
F, and E are points used to construct the nine point circle for
both triangle ABC and triangle HBC.

By similar reasoning it can be proved
that triangle HAB and triangle HAC also share this same nine point
circle.

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