Assignment 8:

Orthocenters and Circumcircles

by

Jenny Johnson

**What is an orthocenter?**

The
orthocenter of a triangle is the point of concurrency of the three lines
containing the altitudes. An
altitude of the triangle is the line segment from a vertex to the opposite side
and perpendicular to the opposite side.
The orthocenter will not always be inside the triangle. This is why the orthocenter is the
point of concurrency of the lines containing the altitudes rather than just the
altitudes. Also, the line
containing the altitude must be perpendicular to the line containing the
opposite side, not just the opposite side as a segment for the cases of obtuse
angles.

First,
I constructed triangle ABC with an orthocenter of H. I made the triangle acute so that the orthocenter would be
inside the triangle initially.
Since the triangle is acute, the orthocenter lies on the three
altitudes. Then, I constructed the
orthocenters of triangle HBC (E), triangle HAB (F) and triangle HAC (G). The orthocenter of HBC was just the
other vertex of the triangle – A.
The orthocenter of HAB was the vertex C and the orthocenter of HAC was
the vertex B. I dragged the
vertices A, B, and C to form several different types of triangles –
isosceles, right, and obtuse – and the orthocenter E was always equal to
A, orthocenter F was always equal to C and orthocenter G was always equal to
B. Pictures of an acute, right and
obtuse ABC are shown below.

It appears that E = A, F = C and G = B will always be true.

**How do we prove it?**

I drew the lines
containing the altitudes of my original acute triangle ABC. The line containing the altitude from
vertex A goes through H and intersects line BC at a right angle. The line
containing the altitude from vertex B goes through H and intersects line AC at
a right angle. The line containing
the altitude from vertex C goes through H and intersects line AB at a right
angle.

Now letŐs look at triangle AHC. In order to find the orthocenter for this triangle, I would
construct the lines containing the altitudes for each of the vertices A, H and
C. H, B and L are collinear, so
the line BL is perpendicular to line AC, which is the common side to both triangles
ABC and ACH. Thus, the altitude HL
from H is contained in this line already. Does the altitude need to be unique?
LetŐs say there is another line that contains an altitude through H
perpendicular to AC at a point P.
That would form the triangle HLP that has two right angles. This is a contradiction. Thus, there is a unique line that is perpendicular to AC (or any line)
through point H (or any given point).
This line is HB and contains the altitude from H for triangle AHC. The foot of the altitude from H is L.

Now, letŐs look at the line containing the altitude from A
perpendicular to the line containing the opposite side HC. Well, line AB is perpendicular to HC by
construction. Since there is a
unique perpendicular to a line from a given point, then AB must be the unique
perpendicular from A to line CH.
The foot of the altitude from vertex A is M.

Finally, the line containing the altitude from vertex C
perpendicular to the line containing the opposite side HA must be the line CB
since CB is perpendicular to HA by construction. The foot of the altitude from vertex C is U for triangle
ACH. Thus, the three lines containing the altitudes of triangle ACH are HB, AB,
and CB which all intersect at point B.
Point B is therefore the point of concurrency of the lines containing
the altitudes (perpendiculars from the vertices of ACH), and hence B is the
orthocenter of triangle AHC.

We have proven that the orthocenter of triangle AHC will
always lie on the vertex B.
Similarly, we can show that
the orthocenter of AHB will always be C and the orthocenter of CHB will
always be A using the same arguments.

**What is a circumcircle?**

A circumcircle is a triangleŐs circumscribed circle. This means it is the unique circle that
passes through the three vertices of the triangle. It is constructed by finding the point of concurrency of the
triangleŐs three perpendicular bisectors.
This point, called the circumcenter, is the center of the
circumcircle. The radius is the
distance from the circumcenter to any of the vertices.

I
constructed the circumcircles of triangles ABC, AHC, AHB, and CHB. Then, I connected the circumcenters of
the circumcircles for triangles AHC, AHB and CHB to form another triangle. See the picture below.

It appears
that the circumcenter N of the circumcircle for triangle ABC is also the
orthocenter of our newly formed triangle OPQ. I constructed the lines containing the altitudes for
triangle OPQ and they intersected at N.
Triangle OPQ is congruent to triangle ABC with corresponding sides
parallel to each other.

The
four circumcircles also appear to be congruent. Now letŐs construct the 9-point circle for each of our four
triangles. The 9-point circle ends
up being identical for all four triangles and is the circle outlined in the
picture below.

**Why is the 9-point circle identical for
all four triangles?**

** **

Well,
for triangle ABC, the 9-point circle goes through the feet of the three
altitudes L, M and U. Three points
determine a circle. Then the 9-point
circle for triangle AHC has to go through the feet of its three altitudes,
which are again L, M and U.
Similarly, for triangles AHB and CHB, the feet of their altitudes will
be L, M and U and therefore they will have the same 9-point circle determined
by these three points. Thus, the
9-point circle is the same for all four triangles.

**Why are the circumcircles equal?**

Now
consider the triangle CHN (seen below by a dashed blue line). We know that E is the midpoint of NH
since the center of the 9-point circle is the midpoint of the segment
connecting the orthocenter with the circumcenter. Also, point V is
the midpoint of CH since the 9-point circle goes through the midpoint of the
segments connecting the orthocenter with the vertices of the triangle.
So, VE is a midsegment for triangle CHN making it parallel to CN and half
the length of CN. Thus, the radius of the 9-point circle is half the
radius of the circumcircle. Since the 9-point circle is the same for all
triangles, then the circumcircles must be congruent for all four triangles as
well.

Do your own exploration
of orthocenter and circumcircles here.