The Eternal TriangleAdam McBride
Based on a presentation at the SMC Conference, University of Stirling, 28 April 2001
Introduction
The story starts one lunchtime in August 1998. A group of bright young postgraduates are perusing the problems set in the 1998 International Mathematical Olympiad which I had brought back from Taiwan. The problems are tough. (You can see for yourself in my article in SMC Journal 28, pages 13-15.) I expected the students to run into trouble sooner or later but I had hoped that they would get a little further than the first problem. Among all these experts in things like fluid mechanics and numerical analysis, none knew what a cyclic quadrilateral was. (However, there were some novel suggestions!)
Although this does not quite represent the end of civilisation, it is nevertheless a poor show and epitomises the lack of basic geometrical knowledge among the vast majority of today's undergraduates in science and engineering. Forty years ago when I was a lad, life was rather different. Geometrical deductions were part of the curriculum for a large percentage of each cohort. This percentage was almost certainly too large, with many pupils finding the going very hard indeed. The 'solution' adopted was to do away with almost everything and introduce new approaches, with topics like transformation geometry and tessellations becoming flavour of the month.
One major casualty in this revolution was the idea of proof in
mathematics. Geometry provided an excellent vehicle for developing powers of
deductive reasoning, which are needed in all branches of mathematics.
Undergraduates embarking on a first course in analysis are hit by a double
whammy. The dreaded 
and 
are bad enough but, in addition to learning what is tantamount to a foreign
language, students have to string the 'words' together to form sentences and
logical arguments. A 'proof' produced nowadays by a typical student usually
contains most of the right words but all in the wrong order.
In my view (and all views in this article are personal), the prospective scientists and engineers are being short-changed by the way geometry is taught at present. At the other end of the spectrum are lots of people who will get through life perfectly happily without knowing that the altitudes of a triangle are concurrent. How to satisfy the needs of a disparate population is a challenge that needs to be faced.
Let's go back to the postgraduates eating their sandwiches in 1998. I decided there and then to give them a course of lectures on the geometry of the triangle. The response was remarkable. The topic was totally different from the usual material of such courses and they loved it! At one level they were exposed to sustained logical arguments. At another level they could appreciate the elegance of many of the results.
We have no right to expect any of these flukes and yet they are all nestling inside the humble triangle.
Come with me as we attempt to unravel the mysteries of what's going on. For some readers this will be a voyage of discovery, while for others it may be more of a trip Down Memory Lane. The material contains opportunities for the lowest attainers to draw pretty pictures as well as opportunities to teach high attainers how to construct rigorous proofs. For everyone, there is the chance to wallow in the sheer beauty of it all.
Congruent Triangles
Any triangle has three sides and three angles. However, we do not need all six bits of information to determine a triangle uniquely. For example, suppose we know that
Draw a line segment of length 2 cm to represent 
. Form an angle of
and head off in
this direction for a distance of 3 cm to arrive at 
. The three vertices
and 
are now known and
the triangle is completely specified.
Figure 1: triangles given SAS
If the three triangles shown in Figure 1 were made of cardboard,
any one could be placed exactly on top of either of the others. In this sense
the triangles are regarded as equivalent or congruent. Knowing the
lengths of the sides 
and 
,
as well as the size of the included angle 
, we can calculate
the length of 
and the size of the angles at 
and 
. Here are four ways of proving two triangles congruent.
SAS was the one discussed above. To illustrate AAS consider
and 
. Note that 
and 
correspond to 
and 
respectively.
Hence 
and 
are corresponding
sides. We say that 
is congruent to 
and write 
.
(Note that we use 
and not 
because 
correspond to 
respectively in Figure 2.)
Figure 2: triangles given AAS
In contrast, knowing the three angles of a triangle is not enough to specify a triangle uniquely (although it would be unique up to similarity, rather than congruence). Likewise two pairs of angles and a pair of non-corresponding sides are not sufficient.
In days of yore, before the advent of vectors, the proofs in school geometry books were based fairly and squarely on the use of the four cases of congruence listed above. We are talking here of books like those by Walker and Millar (12s 3d in old money) or White and Morrison. It is instructive to recall what such proofs looked like. (I'll only include two in detail now, but those of a frail disposition may wish to jump ahead.) Let us consider the (internal) bisectors of the angles of a triangle. As a starter for ten, we need
Theorem 1 The locus of a point equidistant from two intersecting straight lines is the bisector of the angle between them.
There is actually quite a lot to be done here as we shall now see.
Figure 3: 
equidistant from 
and 
Part 1 We shall prove that any point which is equidistant from two intersecting lines lies on the bisector of the angle between them. Let us set things up in Figure 3.
Given:- Straight lines 
and 
meeting at 
;
any point equidistant from 
and 
.
RTP (Required to prove):- 
lies on the bisector of 
.
Construction:- Join 
. Draw 
perpendicular to 
respectively.
Proof:- In 
and 
,
1. 
(given; 
is equidistant from 
and 
)
2. 
(
is
a common side of the two triangles)
3. 
(right angles, by construction).
\ 
(RHS).
\ 
.
\ 
is on the bisector
of 
.
Don't relax yet. We are only half finished!
Part 2
Figure 4: 
on the bisector of 
Given:- Straight lines 
and 
meeting at 
,
any point on the bisector of 
.
RTP:- 
is equidistant from 
and 
.
Construction:- Draw 
, 
perpendicular to 
respectively as in Figure 4.
We want to prove that 
.
Proof:- In 
and 
1. 
(given)
2. 
(right angles, by construction)
3. 
(common side)
\ 
(AAS; 
is opposite the
right angle in both triangles).
\ 
.
\ 
is equidistant from
and 
.
Only now can we add Q.E.D. (quod erat demonstrandum) if we wish!
Theorem 2 The internal bisectors of the three angles of a triangle are concurrent. The point of concurrence is called the incentre and is the centre of the inscribed circle which touches the three sides of the triangle.
Figure 5: concurrence of internal bisectors
Given:- 
.
RTP:- The bisectors of 
, 
and 
are concurrent.
Construction:- Draw 
and 
, the bisectors of 
and 
, respectively which meet at 
, as shown in Figure 5.
Draw 
perpendicular to 
respectively.
Join 
.
We shall show that 
bisects 
.
Proof:- 
is the bisector of 
.
\ 
is the locus of
points equidistant from 
and 
(Theorem 1, Part 2).
\ 
.
Similarly 
.
\ 
.
\ 
is equidistant from
and 
.
\ 
lies on the
bisector of 
(Theorem 1, Part 1).
Also, since 
, 
is the centre of a circle which passes through 
and therefore
touches the three sides of 
.
Q.E.D.
Commentary
The old style was nothing if not disciplined. In order we have
Note the use of '\' for 'therefore' rather than the
all-pervasive 'Û' which is frequently misused and abused nowadays. Theorem 1 also
highlights the difference between a statement and its converse (i.e. 
versus 
).
Note Those of a frail disposition can rejoin us here!
Triangle Centres
There are four well-known (?) sets of three lines, associated
with a triangle 
, which meet at a point.
.
is the centre of the circumcircle of 
, i.e. the unique
circle which passes through the three vertices of the triangle.
The letters 
, 
, 
and
will be used
for these special points from now on.
As mentioned earlier, we have no right to expect three lines to be concurrent. Yet here we have four such 'flukes'. However, these are not all independent of each other.
Theorem 3 Given that the perpendicular bisectors of the sides of any triangle are concurrent, it follows that the altitudes of any triangle are concurrent.
Proof:- We shall adopt a streamlined version of the style used in Theorem 1 and 2.
Let 
be any triangle. We wish to prove that its altitudes are
concurrent.
Through each vertex draw a line parallel to the opposite side
(i.e. lines through 
parallel to 
respectively). Let these intersect at 
as shown in Figure
6. It is not hard to prove that 
and 
are parallelograms so that 
, both being equal to 
.
Figure 6: altitude of 
Let 
be the altitude of 
through 
. Since 
and 
are parallel, 
is perpendicular to 
. From above, it follows that 
is the
perpendicular bisector of 
. By a similar argument applied to the other two altitudes
of 
, we see
that the altitudes of 
are the perpendicular bisectors of the sides of 
. By assumption,
the perpendicular bisectors of 
are concurrent. Hence the altitudes of 
are
concurrent.
Q.E.D.
Here is another interconnection.
Definition Let 
be the feet of the altitudes from 
respectively in 
as shown in Figure 7. Then 
is called the pedal triangle of 
.
Figure 7: pedal triangle of 
Figure 7 contains a lot of rich structure. At this
stage, the dreaded cyclic quadrilateral surfaces. (For younger readers
in the same boat as the postgraduates, a cyclic quadrilateral is a
quadrilateral whose four vertices all lie on a circle.) In the diagram,
is the orthocentre of 
. We know that
\ 
is a cyclic quadrilateral (opposite angles supplementary).
\ 
(angles in same segment of circle 
).
Similarly 
since 
is a cyclic quadrilateral.
Also 
since 
is a cyclic quadrilateral.
\ 
(relabelling angles).
\ 
.
\ 
bisects 
.
We have therefore proved the following result.
Theorem 4 The altitudes of a triangle are the angle bisectors of the corresponding pedal triangle.
Euler Line
Next let us investigate connections between various triangle centres. To get the full flavour, we shall work with a scalene
triangle, i.e. one in which the sides all have different lengths. (If
the original triangle is isosceles or equilateral, diagrams will
degenerate. For example, if 
is equilateral, 
, 
, 
and 
coincide.)
Use of some package such as Geometer's Sketchpad or Cabri Geomètre may suggest various results. For example, it looks as though 
, 
and 
lie on a line and that 
trisects 
. Can this be true?
Theorem 5 The centroid 
, the orthocentre 
and the circumcentre 
of a (scalene) triangle lie on a straight line, called the Euler Line of the triangle. Further, in vector notation,
Proof:-
Figure 8: Euler line of 
Let A' be the mid-point of 
, so that 
is the perpendicular bisector of 
and 
is the median through 
.
It is a standard result that 
.
Let 
be the point on the line through 
and 
such that 
, as shown in Figure 8.
We shall prove that 
is actually 
.
In 
and 
1. 
(from above)
2. 
(vertically opposite angles).
\ 
and 
are similar.
\ 
.
\ 
is parallel to 
.
\ 
(extended) is perpendicular to 
(since 
).
\ 
lies on the altitude of 
through 
.
Similarly 
lies on the altitudes of 
through 
and 
.
\ 
is the orthocentre 
of 
.
\ 
are collinear and 
.
Q.E.D.
Theorem 5 provides us with another wonderful surprise and one which escaped all the Greek geometers. (Euler discovered this in the 18th century.) We shall meet the Euler line again in a moment.
Meanwhile, let us turn our attention to circles related to a triangle. We have already met the incircle and the circumcircle, but that's only a start.
The 9-point circle
Given that three non-collinear points determine a circle uniquely, the likelihood of nine special points lying on a circle seems remote indeed. However, we are in for another wonderful surprise.
Theorem 6 Given 
, there is a circle passing through the following nine points:
the mid-points of 
, 
, 
respectively
the feet of the altitudes 
, 
, 
respectively
the mid-points of 
, 
, 
respectively.
Further, the centre 
of the circle is the mid-point of 
and the radius of the circle is 
, where 
is the radius of the circumcircle of 
shown in Figure 9.
Figure 9: 9-point circle
Proof:- To start with we shall use vectors and take 
, the circumcentre, as origin (a shrewd choice!). Let 
, 
, 
.
By definition of the circumcircle,
By a standard result 
.
\ 
(Theorem 5).
Also 
(
is the mid-point of 
).
Now 
(
is the mid-point of 
).
\ 
.
\ 
(by(*)).
Similarly 
.
\ 
and 
lie on the circle with centre 
and radius 
.
Next 
.
\ 
.
\ 
.
Similarly 
.
\ 
and 
lie on the circle with centre 
and radius 
.
From above 
and 
.
\ 
is the mid-point of 
.
However 
subtends a right-angle at 
.
\ 
is a diameter of the circumcircle of 
and
is the centre of this circle.
\ 
.
Similarly 
.
\ 
and 
lie on the circle with centre 
and radius 
.
Q.E.D.
For your further amazement, we can state
Theorem 7 The 9-point circle of 
is also the 9-point circle of each of 
, 
and 
.
Proof:- Consider 
in Figure 9. It is enough to prove that the 9-point circle of 
is the unique circle passing through the mid-points of the sides 
, 
and 
. However, these are the points, 
and 
in the notation of Theorem 6 and do indeed lie on this circle.
The same applies to 
and 
.
Q.E.D.
Three more circles now come into the picture. We consider one in detail, with the other two being similar. We shall bisect 
of 
internally as before. However, we now bisect 
and 
externally as shown in Figure 10.
Figure 10 : escribed circle opposite 
Theorem 8 The internal bisector of 
and the external bisectors of 
and 
are concurrent at a point 
, which is the centre of a circle which touches 
internally and the sides 
and 
externally (in the sense that 
and 
are extended beyond 
and 
).
Definition In the notation of Theorem 8, 
is called the excentre of 
opposite 
and the corresponding circle is called the escribed circle opposite 
.
There are excentres and escribed circles opposite vertices 
and 
as well.
Relations between the various circles
We now have 6 circles and it is of interest to see how they interact with each other. The following table is useful.
| Circle | Centre | Radius |
| Circumcircle | 
| 
|
| Inscribed | 
| 
|
Escribed, opposite 
| 
|  |
Escribed, opposite 
| 
|  |
Escribed, opposite 
| 
| 
|
| Nine-point | 
| 
|
Table 1
Calculation of the radii of the inscribed and escribed circles involves the use of trigonometric identities and is left as an exercise for the reader. Here are two more such exercises.
, so that 
.
However, rather than dwell on algebraic manipulation, we turn to yet another remarkable result, proved by Karl Feuerbach in 1822.
Theorem 9 (Feuerbach's Theorem)
The 9-point circle of a triangle touches the inscribed circle and each of the three escribed circles.
What right have we to expect this? None at all. We shall not give a proof but instead invite the reader to get a sheet of A3 paper and draw a large diagram.
Challenge Starting with a large scalene triangle, construct every special point and circle mentioned so far, using only a straight edge (with no scale) and a pair of compasses. If you succeed, frame the drawing and hang it on the wall in your classroom.
While we are in this neck of the woods, look at the diagram below in which the angles of an arbitrary triangle 
have been trisected. What do you think is true of 
?
Figure 11: Morley's theorem
Theorem 10 (Morley's Theorem; 1899)
(i) 
is equilateral.
(ii) Each side of 
has length 
, where 
is the circumradius of 
.
The formula for the side-length is reminiscent of those in Table 1. However, there is one new feature. You can bisect an angle using a straight edge and compasses but you cannot trisect an angle in this way. The reason must wait for another time.
Conclusion
We have explored a number of remarkable surprises which
we have no right to expect. Actually this is only the tip of the
iceberg. For example, look again at Theorem 9. The 9-point circle of 
contains 4 more interesting points, where it touches the inscribed and escribed circles of 
. Moreover, Theorem 7 tells us that we'll get another 4 interesting points from each of 
, 
and 
. So our 9-point circle now has 25 interesting points lying on it!
As for interesting points inside a triangle, 
and 
are a mere drop in the ocean. Find hundreds more 'triangle centres' at the website
http://cedar.evansville.edu/~ck6
which is lovingly maintained by Clark Kimberling.
The next time you see a triangle, just think of all that's going on inside it!