Some students in my geometry class were strugglinig with the concepts of cross ratio and harmonic division, so I went looking for some problems to work for them as examples of using these concepts. In Howard Eves' book, College Geometry, I found the following problem:
If are the feet of the altitudes on the sides of a triangle , show that
This sounded like a nice problem, but when I worked it in preparation
for class, I discovered that I did not use the fact that
were altitudes. The only property of altitudes that I used was the
fact that the altitudes of a triangle are concurrent. Therefore my
solution was valid for any concurrent cevian lines
fact I found two nice solutions. One based on the properties of
Ceva's and Menelaus' theorems and the other using the properties of a
After I completed these solutions, I looked at Eves' hint for the
problem to see why he required the cevian lines to be altitudes. His
hint was for the student to use the fact that the altitude
bisected the angle . This certainly gave a solution, except for
one minor point: nowhere in the book (that I could find) was this
fact, or material relating to it, presented. I have a policy, in both
my geometry course and my number theory course, that if a student uses
a hint given by the author of the text, then the student must first
prove the hint is valid before they can use it in their solution. I
decided to see what was involved in writing up a proof for this fact
that bisects angle . When I started working on this, I found
that I was awakening distant memories about properties of the orthic
If AP, BQ, CR are the altitudes for a triangle ABC,
the triangle formed by joining the feet of the altitudes P, Q,
R, is called the orthic triangle for triangle
ABC. (NOTE. Some call this the pedal triangle; however,
in general, a pedal triangle for an acute triangle is the triangle
formed by the feet of the projections of an interior point of the
triangle onto the three sides. Hence there are an infinite number of
pedal triangles for a given triangle.) When looking at the orthic
triangle for a given triangle there are two cases to be
considered: triangle is an acute triangle and triangle is
an obtuse triangle.
I will refer to triangle as the parent triangle. If the parent
triangle is acute, then the altitudes of this triangle bisect the
angles of its orthic triangle; however, if the parent triangle is
obtuse, the angles of the orthic triangle are bisected by the two
sides forming the obtuse angle and the altitude to the side opposite
the obtuse angle. We will look at each case separately.
After this examination, we will show that the orthic triangle is the triangle with minimum perimeter for all triangles inscribe in a given triangle.
Theorem. The altitudes AP, BQ, CR of an acute triangle ABC bisect the angles of its orthic triangle PQR.
Proof. We will show that AP bisects and the others will follow by similar arguments.
We begin by constructing circles
ADDENDUM. From the above discussion, we have some bonus properties. Since the altitudes of a triangle are concurrent and are the internal angle bisectors of the angles of the orthic triangle, we see that the orthocenter is the incenter for triangle . If the altitudes of a triangle are the internal angle bisectors of the angles of the orthic triangle, then the sides are the external angle bisectors since the internal and external bisectors of an angle are perpendicular. We therefore have that the vertices are the centers for the excircles for the orthic triangle. This is illustrated in Figure 4. I vaguely remember mention of the orthic triangle when I was an undergraduate student but at that time I believe we only talked about the orthic triangle for an acute parent triangle. This got me curious about the case for the obtuse parent triangle. We examne this next.
Theorem. Let be an obtuse triangle, angle the obtuse angle and let , , be the altitudes to the sides
. Let triangle be the orthic triangle for triangle . Then side , altitude and side bisect the angles of the orthic triangle. Furthermore, is the incenter for the orthic triangle.
Proof. In Figure 5, is the given triangle and triangle is its orthic triangle.
We need to prove that , and . To this end we construct circles on sides as diameters.
In Figure 6, we see that since they both subtend arc on circle . We also have that since the subtend arc on circle .Hence
We next consider altitude and angle . In Figure 7, since they both subtend arc on circle . In the middle part of Figure 3 we see that since angles and subtend arc on circle . Now in the right-hand part of Figure 7, we see that since angles and subtend arc on circle . Thus so that and altitude bisects angle .
To finish, we need to show that side bisects angle . In Figure 8 we see that since angles and subtend arc on circle . Also since they subtend arc on circle . Hence and , which imples that bisects angle .
Since the angle bisectors all meet in , it is clear that is the incenter for the orthic triangle .
As a final note, we see in Figure 10 that the vertices and and the orthocenter for triangle are the centers for the excircles of the orthic triangle .
Proof. Consider triangle inscribed in an acute triangle . Reflect across to get and reflect across to get . Since triangles and are isosceles,
and . Hence the polygonal path is equal to the perimeter of triangle . Now if we leave fixed and let and move along and until they lie on line (Figure 2), then the perimeter of triangle will equal .
Now since triangles and are isosceles, . Furthermore, , since is the bisector of and is the bisector of .
Now with and on line , if we now allow to move along , will remain constant with measure the same as . Triangle will always remain an isosceles triangle; however, its base will vary in direct proportion to . Thus will be a minimum when is a minimum and that occurs when is the altitude of triangle to side .
This process can be repeated for and to show that they are the feet of the altitudes from and . Therefore the triangle with minimum perimeter is the orthic triangle. See Figue 13.