My Mathematics dissertation completed in my third year at Coventry University. The topic is on the geometry of triangles and the theorems relating to them.

1. The Secret Life of Triangles
Christopher Thomson
ID: 4265645
Supervisor: Trevor Hawkes
A report presented in the Department of Mathematics and Physics, Coventry University, for the
degree of Bachelor of Science in Mathematics, April 2015.

2. Abstract
This report explores various theorems and properties related to the Euclidean geometry of triangles,
as well as providing proofs and figures to further enhance the explanation. This report contains
four chapters, each with a different key focus.
The first chapter, entitled Three Lines Through a Point, explains and proves theorems relating to
three lines that pass through a point (concurrent), with a particular emphasis on so called ’triangle
centres’, which are listed in Kimberling’s Encylopedia of Triangle Centres. These include, amongst
others, the incentre, the centroid, the orthocentre and the circumcentre.
The second chapter, called Three Points Through a Line, explains and proves four theorems
that involve a line passing through three points (collinear). The first, Menelaus’ Theorem, is used
to prove Desargues’ and Pappus’s Theorems.
The third chapter, entitled Unexpected Equilateral Triangles, gives detail on two theorems that
involve three points forming equilateral triangles. This includes Morley’s Theorem with an extensive
proof, and Napoleon’s Theorem.
The final chapter talks about the nine-point circle and the Euler line, as well as properties and
theorems relating to them.
Each chapter gives detailed figures to provide a visual representation of the theorems. The
majority of these figures have been drawn in GeoGebra. Each section is also cited, with references
listed in the appendices at the end of the report.
Page: 1

3. Contents
Introduction 4
Literature Review 5
1 Three Lines Through A Point 7
1.1 Ceva’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 The Centre of a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Trilinear Co-ordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.4 The Incentre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.5 The Centroid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.6 The Orthocentre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.7 The Circumcentre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.8 The Napoleon Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.9 The Exeter Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2 Three Points On A Line 22
2.1 Menelaus’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 Desargues’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.3 Pappus’s Hexagon Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.4 The Simson Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3 Unexpected Equilateral Triangles 29
3.1 Morley’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2 Napoleon’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2

4. 4 The Nine-Point Circle and The Euler Line 37
4.1 The Nine-Point Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4.2 The Euler Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Conclusion 43
References 44
Appendices 46
Page: 3

5. Introduction
What is a triangle? The simple definition of a triangle is a polygon with three sides and three
vertices. However, there is a lot more to a triangle than that. For example, the internal angles add
up to 180 degrees, and the external angles add up to 360 degrees; a triangle with three equal sides
and three equal angles is called an equilateral triangle; and in a right-angled triangle, the square
of the hypotenuse equals the sum of squares of the other sides (Pythagoras’s Theorem). There
are many well-known properties and theorems relating to triangles, however there are many more
less-known and complex properties and theorems that we can explore.
For example, where is the centre of a triangle? How do you find it, and is there a single centre?
This report aims to answer this question, and many more, in order to discover the secrets of triangles.
In four themed chapters, this report will provide detailed explanations of various properties and
theorems relating to the Euclidean geometry of triangles that have been discovered throughout
history, as well as providing mathematical proofs and detailed diagrams for the majority, to further
enhance the explanation. I will also research into the history of the person responsible for the
discovery and/or proof of the mentioned theorem.
To complete this report, I will undertake research from various sources, gathering information
about a wide range of theorems related to triangles. In the first chapter we will answer the question
about the centre of a triangle, detailing a few possible candidates. In the second chapter we will
focus on theorems that involve three collinear points. In the third chapter we will discuss theorems
that involve three points making an equilateral triangle. And in the final chapter we will discuss
further findings that do not fit into the themes of the previous chapters.
As detailed further in the literature review, I will use a variety of sources for my research, ranging
from books to on-line articles, each with their own strengths and weaknesses. Each source provides
a different use for my research. For example, some sources provide mathematical proofs for stated
theorems, whilst others do not. I hope you will now read on with interest and find this report very
informative!
4

6. Literature Review
This report involved researching various topics relating to the Euclidean geometry of triangles. As
this is a purely desk-based project, the content of this report is based on 100% research. I researched
various properties, theorems and proofs relating to triangle geometry. I have found my literature
through the library and internet searching and chosen them based on the specific topics they cover
as well as the quality of explanations for each. This literature review examines the content of the
main literature sources I used, how I used them, and provides comparisons between each based on
how useful they were to my research.
The first book I used for my research was Geometry Revisited by Coxeter, H.S.M. and Greitzer,
S.L. (1967). This book explains, in detail, a wide range of topics relating to Euclidean geometry that
was beneficial to my research. The first chapter is purely based on geometry relating to triangles,
and was my main area of reading. The third chapter is about collinearity and concurrency, which
also proved useful for my report. The book introduces each topic with a brief history of the person
who discovered and/or proved the theorem that is then mentioned, followed by a full proof of the
theorem. There are also plenty of figures/images that give a visual representation of the theorems.
Another book I used for my report was Exploring Advanced Euclidean Geometry with GeoGebra
by Venema, G.A. (2013). Like Coxeter and Greitzer (1967), this book covers a wide range of topics
relating to Euclidean geometry. There was a larger emphasis on triangle geometry, with seven of
the fourteen chapters related to triangles. Compared to Coxeter and Greitzer (1967), each section
starts with a generally more extensive summary of the history of the person responsible for each
theorem. However, the book often does not provide a proof for each theorem, and the majority of
the time it leaves proving a theorem as a task for the reader in the form of exercises. This means I
had to use multiple sources for a topic to research the theorem and the proof. Each section includes
a figure/image alongside the theorem, often in much better detail than Coxeter and Greitzer (1967).
An on-line resource that contains a high amount of useful information is Wolfram MathWorld,
an extensive mathematical resource created by Eric Weisstein (1995), as well as many other contrib-
utors. This website contains several pages on almost every area of mathematics, including geometry,
which was needed for this report. Each page about a specific topic contains a brief description, as
well as many properties and information relating to the topic. There are also helpful figures/images
to give a visual representation to support my findings. Unlike Coxeter and Greitzer (1967), there
are no proofs of any theorems that are included, so this resource was only used for information and
not proofs. There is also little to no information on the history of the theorem or person responsible
for discovering it, so other sources had to be used for this information.
As well as Wolfram Mathworld, I used many more on-line resources for my research, mostly for
5

7. single topics in this report. For example, I used the website MacTutor and Encylopaedia Brittanica
to research the history on some mathematicians. I also researched various on-line webpages, articles,
PDFs and extracts to find information on specific topics.
All resources I used give an image/figure to provide a visual representation of the topic mentioned
in the section or article. However, I decided, in most cases, to recreate the figures in GeoGebra
myself, to demonstrate that I have understood the topic. GeoGebra is an interactive geometry,
graphing and algebra application, where the user can freely draw geometrical figures. I learnt how
to use this software through practice, as well as the GeoGebra Wiki and various on-line help forums.
I believe I have used a wide range of resources for my research, and have utilised the best
elements from each.
Page: 6

8. Chapter 1
Three Lines Through A Point
In geometry, if three or more lines intersect each other at a single point, they are said to be
concurrent. A single triangle has many ways of drawing three lines that are concurrent, and in most
cases, this point of intersection is called a centre. Before looking further into these centres, we will
first look at a very important theorem.
1.1 Ceva’s Theorem
Giovanni Ceva (c.1647 - c.1734) was an Italian mathematician and professor at the University of
Pisa until becoming the Professor of Mathematics at the University of Mantua in 1686. His main
area of study was geometry, and published many works throughout his life, including Opuscula
mathematica and Geometria Motus (O’Connor and Robertson 2012). However, in his work
De lineis rectis, he published an important theorem:
Theorem 1.1.1 (Ceva’s Theorem). Let ABC be a triangle and D, E and F be points on lines BC,
AC and AB respectively. The lines AD, BE and CF are concurrent iff (if and only if)
BD
DC
·
CE
EA
·
AF
FB
= 1 (1.1)
(Coxeter and Greitzer, 1967: p4-7)
Any line that passes through the vertex of a triangle and the opposite side is called a cevian.
This theorem is important for proving that three cevians are concurrent, and will be used extensively
in this chapter to prove special sets of cevians are concurrent.
Proof. First we show that equation (1.1) is necessary for the three cevians to be concurrent. Assume
they meet at a point G. Let α = BD/DC. We will show that α = ABG/CAG. Draw altitudes
h1 perpendicular from BC to A and h2 perpendicular from BC to G (See Figure 1.1). The area of
ABD is 0.5h1BD and the area of ACD is 0.5h1DC. If we divide ABD and ACD we get
ABD
ACD
=
0.5h1BD
0.5h1DC
7

9. Figure 1.1: Ceva’s Theorem Proof
Cancelling out 0.5h1 from the fraction on the right hand sides leaves
ABD
ACD
=
BD
DC
Now do the same using altitude h2
BGD
CGD
=
0.5h2BD
0.5h2DC
=
BD
DC
Therefore
BD
DC
=
ABD
ACD
=
BGD
CGD
=
ABD − ACD
BGD − CGD
Leaving
α =
BD
DC
=
ABG
CAG
(1.2)
Using the same process, we can show that
β =
CE
EA
=
BCG
ABG
(1.3)
and
γ =
AF
FB
=
CAG
BCG
(1.4)
Multiplying equations (1.2), (1.3) and (1.4) together, we have
αβγ =
BD
DC
·
CE
EA
·
AF
FB
=
ABG
CAG
·
BCG
ABG
·
CAG
BCG
By cancelling terms on the right hand side we obtain
BD
DC
·
CE
EA
·
AF
FB
= 1 (1.5)
Page: 8

10. as desired.
We can also prove the inverse of the theorem to show that (1.1) is sufficient. Assume the cevians
AD and BE intersect at point G, and the third cevian that passes through this point is CF . Then
BD
DC
·
CE
EA
·
AF
F B
= 1
But from theorem 1.1.1 we know that
BD
DC
·
CE
EA
·
AF
FB
= 1
From this we can show that
AF
F B
=
AF
FB
So therefore F and F are the same point, therefore AD, BE and CF are concurrent. QED
(Wilson, n.d.)
(Coxeter and Greitzer, 1967: p4-7)
1.2 The Centre of a Triangle
Where is the centre of a triangle? There is no single way to find the centre, thus there is no single
point that can be considered the centre of a triangle. Some of the candidates for the centre of a
triangle are derived from various geometric ideas. The most common process is by constructing
three lines in the same way from each vertex and/or edge of the triangle, and by using Ceva’s
Theorem we can prove the three lines are concurrent. This single point becomes one of the many
centres of the triangle.
Clark Kimberling (1942-), a Mathematics professor at the University of Evansville, USA, main-
tains an extensive list of triangle centres that have been discovered and proved by many other
mathematicians over the past few millennia. Currently, the list contains over 7,500 triangle centres,
with more being discovered over time. Listed below are a few select centres, most of which will be
discussed further in this report:
Page: 9

11. Ref Sym Centre Definition
X1 I Incentre Intersection of the three angle bisectors, and the centre of
the incircle (Chapter 1.4)
X2 G Centroid Intersection of the three medians (Chapter 1.5)
X3 O Circumcentre Intersection of the three perpendicular bisectors, and the
centre of the circumcircle (Chapter 1.7)
X4 H Orthocentre Intersection of the three altitudes (Chapter 1.6)
X5 N Nine-Point Centre Centre of the nine-point circle (Chapter 4.1)
X6 K Symmedian Point Intersection of the three symmedians (reflection of medians
over corresponding angle bisectors)
X7 Ge Gergonne Point Symmedian point of contact triangle (made from points
where incircle meets triangle)
X8 Na Nagel Point Intersection of lines from each vertex to the corresponding
semiperimetre point
X9 M Mittenpunkt Symmedian Point of the triangle formed by the centres of
the three excircles
X10 Sp Spieker Centre Centre of the Spieker circle (incircle of the medial triangle)
X11 F Feuerbach Point Point where incircle is tangent to the nine-point circle
(Chapter 4.1)
X13 F Fermat Point Point where total distance from the three vertices is mini-
mum
X17 N1 First Napoleon Point Intersection of the lines from vertices to centroids of exter-
nally drawn equilateral triangles (Chapter 1.8)
X18 N2 Second Napoleon Point Intersection of the lines from vertices to centroids of inter-
nally drawn equilateral triangles (Chapter 1.8)
X20 L de Longchamps Point Reflection of orthocentre around the circumcentre
X21 S Schiffler Point Concurrence of the Euler Lines of ABC, IBC, IAC and
IAB, where I is the incentre (Chapter 4.2)
X22 Ex Exeter Point (See Chapter 1.9)
(Kimberling, c.1994)
Figure 1.2 gives a representation of the many triangle centres that have been discovered. How-
ever, it does not show all of them, as only so many can be shown on one image.
1.3 Trilinear Co-ordinates
If we have a triangle ABC, we may need to know where a particular point, P say, in the same
plane is. This position is determined by the ratios of the distances from P to the sides of triangle
ABC (See Figure 1.3). These ratios are the trilinear co-ordinates of P and are written in the form
α : β : γ, where α, β and γ are the directed ratio of the distances from X to BC, AC and AB
respectively. For example, the vertices of triangle ABC have co-ordinates A = 1 : 0 : 0, B = 0 : 1 : 0
and C = 0 : 0 : 1. (Kimberling, 1994)
Page: 10

12. Figure 1.2: Triangle Centres (Weisstein, ”Kimberling Center”)
Figure 1.3: Trilinear Co-ordinates
Page: 11

13. 1.4 The Incentre
Definition 1.4.1 (Internal Angle Bisector). A line that passes through a vertex, so as to cut the
vertex in half internally.
Theorem 1.4.2 (Angle Bisector Theorem). The three internal angle bisectors of a triangle are
concurrent.
Definition 1.4.3 (Incentre). The point of intersection of the internal angle bisectors is called the
incentre.
The incentre is often labelled as I, or X1. The incentre lies equal distance from each of the
three sides of the triangle. The incentre is also the centre of the incircle, a circle that fits inside
the triangle so that the three sides form tangents to the circle. The distance from each side of the
triangle to the incentre is known as the inradius. (See Figure 1.4) (Coxeter and Greitzer, 1967:
p11-14)
The Cartesian co-ordinates of the incentre can be determined by
(xI, yI) =
axA + bxB + cxC
a + b + c
,
ayA + byB + cyC
a + b + c
where a, b and c are the lengths of sides BC, AC and AB respectively.
The trilinear co-ordinates of the incentre are 1 : 1 : 1, due to it being equidistant from each side
of the triangle. (Weisstein, ”Incenter”)
Figure 1.4: Incentre and Incircle
Page: 12

14. Proof. Let ABC be a triangle, D, E and F be points on BC, AC and AB respectively, and AD,
BE and CF be the angle bisectors of ABC.
Lemma 1.4.4. Let ABC be a triangle and the angle bisector of ∠CAB intersect BC at D. Then
AB
AC
=
BD
CD
Using the above lemma for each fraction in Ceva’s Theorem
BD
DC
=
BA
AC
CE
EA
=
CB
BA
AF
FB
=
AC
CB
This leads to
BD
DC
·
CE
EA
·
AF
FB
=
BA
AC
·
CB
BA
·
AC
CB
= 1
Therefore the three angle bisectors of triangle ABC are concurrent. QED
(Faucette, 2007)
Excircles and Excentres
Excircles, also known as escribed circles, are circles that lie fully outside the triangle. An excircle
is drawn tangent to one side of the triangle, externally, and the two other sides when they are
extended outwards. Each triangle has three distinct excircles, one for each side. The centre of
each excircle is called the excentre. These centres can be found by the intersection of the internal
angle bisectors of the opposite angle, and the external angle bisectors (perpendicular to the internal
angle bisectors) of the other two angles (See Figure 1.5). Excentres are often labelled Ji, where i
is the vertex of ABC opposite the excentre. The trilinear co-ordinates of each of the excentres are
−1 : 1 : 1, 1 : −1 : 1 and 1 : 1 : −1. (Weisstein, ”Excircles”)
1.5 The Centroid
Definition 1.5.1 (Median). A line that passes through a vertex of a triangle and the midpoint of
the opposite side is called a median.
Theorem 1.5.2 (Median Concurrence Theorem). The three medians of a triangle are concurrent.
Definition 1.5.3 (Centroid). The point of intersection of the medians is called the centroid.
Page: 13

15. Figure 1.5: Excircles and excentres
A median effectively cuts the area of a triangle in half, therefore the three medians of a triangle
will cut it into six smaller triangles of equal area. The centroid is often labelled G, or X2. The
centroid of a triangle is also known as the centre of mass. That is, if the triangle was a physical,
2D object, it would balance on this point (See Figure 1.6). (Coxeter and Greitzer, 1967: p7-8)
The Cartesian co-ordinates of the centroid can be determined by
(xG, yG) =
xA + xB + xC
3
,
yA + yB + yC
3
There are three ways to calculate the trilinear co-ordinates of the centroid:
bc : ca : ab = 1/a : 1/b : 1/c = csc(A) : csc(B) : csc(C)
(Weisstein, ”Centroid”)
Proof. Let ABC be a triangle, D, E and F be the midpoints of BC, AC and AB respectively, and
AD, BE and CF be the medians of ABC. The median from vertex A intersects the midpoint, D,
of BC. Therefore BD is equal to DC, and BD/DC = 1. The same follows for the other medians.
So CE/EA = 1 and AF/FB = 1. Substituting these into Ceva’s theorem
BD
DC
·
CE
EA
·
AF
FB
= 1 · 1 · 1 = 1
Therefore the three medians of triangle ABC are concurrent. QED
Page: 14

16. Figure 1.6: Centroid
(Faucette, 2007)
1.6 The Orthocentre
Definition 1.6.1 (Altitude). A line that passes through a vertex of a triangle and the opposite
side, so that the line is perpendicular to the side, is called an altitude.
Theorem 1.6.2 (Altitude Concurrence Theorem). The three altitudes of a triangle are concurrent.
Definition 1.6.3 (Orthocentre). The point of intersection of the altitudes is called the orthocen-
tre.
The orthocentre is often labelled H, or X4. If the triangle is obtuse (contains an angle larger
than 90 degrees), then the orthocentre will lie outside the triangle. When drawing the altitudes,
the two edges either side of the obtuse angle must be extended outside the triangle. If the triangle
is right-angled (contains an angle of 90 degrees), then the orthocentre coincides with the right angle
(See Figure 1.7). (Coxeter and Greitzer, 1967: p9)
The Cartesian co-ordinates of the orthocentre can be determined by
(xH, yH) =
xA tan(A) + xB tan(B) + xC tan(C)
tan(A) + tan(B) + tan(C)
,
yA tan(A) + yB tan(B) + yC tan(C)
tan(A) + tan(B) + tan(C)
where A, B and C are the sizes of the respective angles.
Page: 15

17. The trilinear co-ordinates of the orthocentre are sec(A) : sec(B) : sec(C). (Weisstein, ”Ortho-
center”)
Figure 1.7: Orthocentre
Proof. Let ABC be a triangle, D, E and F be points on BC, AC and AB respectively, and AD,
BE and CF be the altitudes of ABC. Take the right-angled triangles BEC and ADC. Because
∠BCE is acute, ∠BCE = ∠ACD. This angle together with ∠CBE equals 90 degrees, similarly
this angle with ∠CAD equals 90 degrees. Therefore BEC and ADC are similar triangles where
CE/DC = BC/AC. A similar process with triangles FAC and EAB leads to AF/EA = AC/AB,
and with triangles BFC and BDA leads to BD/FB = AB/BC.
BD
DC
·
CE
EA
·
AF
FB
=
CE
DC
·
AF
EA
·
BD
FB
=
BC
AC
·
AC
AB
·
AB
BC
= 1
Therefore the three altitudes of triangle ABC are concurrent. QED
(Faucette, 2007)
1.7 The Circumcentre
Definition 1.7.1 (Perpendicular Bisector). A line that passes through the midpoint of a side of a
triangle, and is perpendicular to the side, is a perpendicular bisector.
Theorem 1.7.2 (Perpendicular Bisector Concurrence Theorem). The three perpendicular bisectors
of a triangle are concurrent.
Page: 16

18. Definition 1.7.3 (Circumcentre). The point of intersection of the perpendicular bisectors is called
the circumcentre.
The circumcentre is often labelled O, or X3. The circumcentre lies equidistant from the three
vertices of the triangle. It is also the centre of the circumcircle of the triangle, which is a circle that
passes through all three vertices (See Figure 1.8). (Coxeter and Greitzer, 1967: p7)
The Cartesian co-ordinates of the circumcentre can be determined by the following process:
D = 2(xA(yB − yC) + xB(yC − yA) + xC(yA − yB))
Ux = ((x2
A + y2
A)(yB − yC) + (x2
B + y2
B)(yC − yA) + (x2
C + y2
C)(yA − yB))/D
Uy = ((x2
A + y2
A)(xC − xB) + (x2
B + y2
B)(xA − xC) + (x2
C + y2
C)(xB − xA))/D
(xO, yO) = (Ux, Uy)
The trilinear co-ordinates of the circumcentre are cos(A) : cos(B) : cos(C). (Weisstein, ”Cir-
cumcenter”)
Figure 1.8: Circumcentre
Proof. Draw the perpendicular bisectors of AC and BC. The point of intersection is at point O.
As any point on the perpendicular bisector of AC is equidistant from A and C
OA = OC
Likewise for the bisector of BC
OB = OC
Page: 17

19. Therefore
OA = OB
Because any point equidistant from the end points of a segment lies on its perpendicular bisector,
O must lie on the perpendicular bisector of AB, proving the existence of the circumcentre.
Since OA = OB = OC, point O is equidistant from A, B and C. Also, this means that there exists
a circle with centre at O that passes through all three vertices of the triangle, therefore proving the
existence of the circumcircle.
QED
1.8 The Napoleon Points
Napoleon points are another type of triangle centres, found from the intersection of three lines.
However these points are found slightly differently. The method is to construct an equilateral
triangle on each of the three sides of triangle ABC, and connect each vertex of ABC to the centre
of the equilateral triangle connected to the opposite side with a line. These lines are concurrent at a
point known as a Napoleon Point. As the equilateral triangles can be drawn outwardly or inwardly,
there are two Napoleon points, called First and Second Napoleon points.
Figure 1.9: First Napoleon Point
To find the First Napoleon point for triangle ABC, outwardly construct equilateral triangles on
Page: 18

20. sides BC, CA and AB, as in Figure 1.9. Label these triangles BCD, ACE and ABF. Let X, Y
and Z be the centres of these triangles respectively. Construct a line that passes through AX, BY
and CZ. These lines are concurrent, and meet at the point N1, known as the first Napoleon point.
Finding the Second Napoleon point, or N2, follows the exact same process, but with constructing
the triangles BCD, ACE and ABF inwardly. (Weisstein, ”Napoleon Points”)
Generalisation
When defining Napoleon points, we always draw equilateral triangles on each sides of the triangle
and consider their centres. We can consider the centres of these equilateral triangles as the vertex
angle of an isosceles triangle, connected to the sides of triangle ABC, and the base angles equal to
30 degrees. A generalisation of the Napoleon points determines which other triangles constructed of
the sides of triangle ABC, have concurrent lines when joining the external vertices and the vertices
of ABC.
If the three triangles XBC, Y CA and ZAB, constructed on each side of triangle ABC, have
equal base angles, θ, then the three lines AX, BY , CZ are concurrent at N (See Figure 1.10). The
trilinear co-ordinates of X, Y , Z and N are:
X = − sin(θ) : sin(C + θ) : sin(B + θ)
Y = sin(C + θ) : − sin(θ) : sin(A + θ)
Z = sin(B + θ) : sin(A + θ) : − sin(θ)
N = csc(A + θ) : csc(B + θ) : csc(C + θ)
(Weisstein, ”Napoleon Points”)
Figure 1.10: Generalisation of the Napoleon Point
Page: 19

21. Proof. The intersections of AX and BC is at A , BY and AC is at B and CZ and AB is at C
(See Figure 1.10). First we find the areas of BCZ and ACZ.
BCZ = BC · BZ · sin(B + θ)
ACZ = AC · AZ · sin(A + θ)
As BZ = AZ
ACZ
BCZ
=
AC · sin(A + θ)
BC · sin(B + θ)
Due to ACZ and BCZ having the same base, CZ, the altitudes from B and C to CZ have the
same ratios as the areas, therefore
AC
BC
=
AC · sin(A + θ)
BC · sin(B + θ)
We can use the same process to find the ratios for BA /CA and AB /CB . Multiply them to get
AC · sin(A + θ)
BC · sin(B + θ)
·
AB · sin(B + θ)
AC · sin(C + θ)
·
BC · sin(C + θ)
AB · sin(A + θ)
= 1
From Ceva’s Theorem, it follows that AA , BB and CC are concurrent. It then follows that AX,
BY and CZ must also be concurrent. QED
(Floor, 1998)
If the three triangles are similar (same sized angles), then the three lines AX, BY , CZ are
concurrent at N.
Furthermore, if the two base angles connected to each vertex of ABC are equal, ie ∠CBX =
∠ABZ, ∠ACY = ∠BCX and ∠BAZ = ∠CAY , then the lines AX, BY , CZ are concurrent at N.
1.9 The Exeter Point
The Exeter point is another special point, and is classed as a triangle centre. To find the Exeter
point of triangle ABC, draw the circumcircle, the circle that meets all three vertices of ABC.
Extend the medians from A, B and C, so they intersect the circumcircle at points A , B and C
respectively. Let DEF be a triangle formed from the tangents of the circumcircle at points A, B
and C, with D being opposite A, E being opposite B and F being opposite C. The lines that
pass through DA , EB and FC are concurrent, and meet at the Exeter Point (See Figure 1.11).
(Weisstein, ”Exeter Point”)
Page: 20

22. Figure 1.11: The Exeter Point
Page: 21

23. Chapter 2
Three Points On A Line
2.1 Menelaus’ Theorem
Menelaus’ theorem dates back to approximately AD 100. Menelaus of Alexandria (c.70-140), a
Greek Mathematician and astronomer, published the treatise Sphaerica, focusing on the geometry
of the sphere. In Book III of this treatise, he introduces a special theorem in spherical geometry that
later became Menelaus’ theorem. He uses this theorem, that we will state and prove, as a lemma
to prove a spherical version. It is unknown who actually discovered the theorem. (Encylopaedia
Brittanica, 2015)
Theorem 2.1.1 (Menelaus’ Theorem). Let ABC be a triangle, and let X, Y and Z be points on
the sides BC, CA and AB respectively. X, Y and Z are collinear iff
BX
XC
·
CY
Y A
·
AZ
ZB
= −1 (2.1)
Equation (2.1) uses signed lengths for each segment. For example, AB will be directed positively
from A to B, and BA = −AB. Orient the sides of the triangle, so that AB, BC, and CA are positive.
Then, for instance, the ratio BX/XC will be positive iff X lies strictly between B and C. As a
result, there will always be either one or three of the points X, Y and Z outside the triangle, the
product on the left of equation (2.1) will always be negative. (Coxeter and Greitzer, 1967: p66-67;
Venema, 2013: p78-80)
Proof. We first prove that the equation (2.1) is a necessary condition for collinearity. Assuming X,
Y and Z are collinear, construct three segments, h1, h2 and h3, perpendicular to the line XZ and
ending at A, B and C respectively, as in Figure 2.1. If these segments are situated on opposite sides
of XZ, treat the lengths on one side as positive and the other as negative. Using similar triangles,
we get the equations
h2
h3
=
BX
XC
(2.2)
h3
h1
=
CY
Y A
(2.3)
22

24. Figure 2.1: Menelaus’ Theorem
h1
h2
=
AZ
ZB
(2.4)
Multiply (2.2), (2.3) and (2.4) to get
h2
h3
·
h3
h1
·
h1
h2
=
BX
XC
·
CY
Y A
·
AZ
ZB
= 1
which shows that X, Y and Z are collinear. Since either one or three of these points lie outside the
triangle, the left-hand side must be negative. Therefore, the equation becomes
BX
XC
·
CY
Y A
·
AZ
ZB
= −1
We can also prove that equation (2.1) is a sufficient condition by proving the converse. We first
assume that equation (2.1) is true. Let X, Y and Z be points such that
BX
CX
·
CY
AY
·
AZ
BZ
= 1
Let AB and XY meet at Z
BX
CX
·
CY
AY
·
AZ
BZ
= 1
Therefore
AZ
BZ
=
AZ
BZ
Z coincides with Z, therefore X, Y and Z are collinear. QED
(Coxeter and Greitzer, 1967: p66-67)
Page: 23

25. 2.2 Desargues’ Theorem
Girard Desargues (1591 - 1661) was a French architect, engineer and mathematician and one of the
founders of projective geometry. He discovered a theorem that has many applications in perspective
drawing. (Field, 1995)
Definition 2.2.1 (Centre of Perspectivity). Two triangles ABC and A B C are perspective cen-
trally if the three lines AA , BB and CC are concurrent at point O. This point is called the
centre of perspectivity.
Definition 2.2.2 (Axis of Perspectivity). Two triangles ABC and A B C are perspective axially
if the intersections of the lines AB and A B (D), lines AC and A C (E) and lines BC and B C
(F) are collinear. The line through D, E and F is called the axis of perspectivity.
Theorem 2.2.3 (Desargues’ Theorem). Iff two triangles are perspective from a single point (cen-
trally), then they are perspective axially, providing the pairs of corresponding sides intersect.
Figure 2.2: Desargues’ Theorem
If any two corresponding sides of the triangles are parallel, then there is no point of intersection.
Therefore Theorem 2.2.3 includes the condition that the pairs of corresponding sides must intersect.
(Coxeter and Greitzer, 1967: p70-72; Venema, 2013: p86-87)
Proof. To prove this theorem, we use Menelaus’ Theorem on the triads of points DB A , EC A and
Page: 24

26. FC B on the sides of the triangles OBA, OAC and OBC respectively. This leads to the equations
BD
DA
·
AA
A O
·
OB
B B
= −1 (2.5)
AE
EC
·
CC
C O
·
OA
A A
= −1 (2.6)
CF
FB
·
BB
B O
·
OC
C C
= −1 (2.7)
Multiply equations (2.5), (2.6) and (2.7) together, and cancel out like terms to get
BD
DA
·
AE
EC
·
CF
FB
= −1
indicating that D, E and F are collinear. QED
(Coxeter and Greitzer, 1967: p70-72)
2.3 Pappus’s Hexagon Theorem
Pappus of Alexandria (c. 290 - 350 AD) is widely considered the last and most important ’geometers
of antiquity’. Pappus’s Hexagon theorem was first proved around AD 300 by Pappus, but its im-
portance in projective geometry wasn’t realised until the sixteenth century. (Coxeter and Greitzer,
1967: p67-70; Venema, 2013: p91-93)
Theorem 2.3.1 (Pappus’s Hexagon Theorem). Let A, B, C, D, E and F be any six points. Let Y
be the intersections of CD and AF, Z be the intersections of BF and CE and X be the intersections
of AE and BD. If A, B and C lie on one line and D, E and F lie on another line, then X, Y and
Z are collinear.
(Venema, 2013: p91-93)
Proof. Let the lines CD, BF and AE form a triangle UV W, as in Figure 2.3. Now use Menelaus’
Theorem on the triads of points AY F, CZE, BXD, ABC and DEF. This leads to the equations
V Y
Y W
·
WF
FU
·
UA
AV
= −1 (2.8)
V C
CW
·
WZ
ZU
·
UE
EV
= −1 (2.9)
V D
DW
·
WB
BU
·
UX
XV
= −1 (2.10)
V C
CW
·
WB
BU
·
UA
AV
= −1 (2.11)
V D
DW
·
WF
FU
·
UE
EV
= −1 (2.12)
Page: 25

27. Figure 2.3: Pappus’s Hexagon Theorem
Take the product of equations (2.8), (2.9) and (2.10), and divide the result by the product of (2.11)
and (2.12) and cancelling gives
V Y
Y W
·
WZ
ZU
·
UX
XV
= −1
Showing that X, Y and Z are collinear, as desired. QED
(Coxeter and Greitzer, 1967: p67-70)
2.4 The Simson Line
Let ABC be any triangle and P a point distinct from A, B and C. Let X be the unique point on
the line AB such that PX and AB are perpendicular. Similarly drop perpendiculars from P onto
BC and AC intersecting Y and Z respectively. These are the closest points on each side of the
triangle to P. The triangle XY Z is known as the pedal triangle and point P is known as the pedal
point.
Theorem 2.4.1 (The Simson Line). If P lies on the circumcircle of triangle ABC, then the three
closest points to P on lines AB, BC and AC are collinear.
The line that is formed from these points is called the pedal line, or the Simson line (See Figure
2.4), named after Robert Simson (1687-1768), a Scottish mathematician. This line can also be
Page: 26

28. described as a degenerate pedal triangle (a triangle with collinear vertices and an area of zero).
(Venema, 2013: p93-96)
Figure 2.4: The Simson Line
Proof. This proof assumes that the pedal point P lies on the circumcircle between B and C. The
same can be proved if P lies between two other points by changing the labels of the points either in
the diagram or in the proof.
Because X, Y and Z are perpendicular, P is a point on the circumcircles of XAZ, XY B and
CY Z. Therefore
∠CPB = 180◦
− A = ∠ZPX
By subtracting ∠CPX, we find that
∠XPB = ∠ZPC
Because points X, B, P and Y lie on a circle
∠XPB = ∠XY B
and because points C, Y , P and Z lie on a circle
∠ZPC = ∠ZY C
Therefore
∠XY B = ∠ZY C
so the points X, Y and Z are collinear. QED
Page: 27

29. (Coxeter and Greitzer, 1967: p40-41)
There are many interesting properties that relate to the Simson Line. First, as you move the
pedal point P around the circumcircle, the Simson Line traces out, or envelopes, a deltoid. This
deltoid is known as the Steiner deltoid.
Also, if you include another pedal point, Q, diametrically opposite P on the circumcircle, and
draw its Simson Line, the two Simson Lines will be perpendicular. Furthermore, if you move the
two pedal points around the circumcircle together, so that they remain diametrically opposite, the
intersection of the Simson Lines will trace out the nine-point-circle (see Chapter 4.1). (Frank and
Weisstein, ”Simson Line”)
Page: 28

30. Chapter 3
Unexpected Equilateral Triangles
3.1 Morley’s Theorem
Frank Morley (1860-1937) was an English mathematician who lived in the USA since 1887. Whilst
he was a professor at the Johns Hopkins University, USA, he edited the American Journal of
Mathematics, as well as several other books. In 1899, Morley discovered a theorem, published in
1929, which is now known as Morley’s Theorem. (Jost and Maor, 2014: p160-163)
Theorem 3.1.1 (Morley’s Theorem). Let ABC be a triangle with internal angles 3α, 3β and 3γ at
vertices A, B and C respectively. Let X be the intersection of the trisectors from A and B adjacent
to side AB. Define Y and Z similarly. Points X, Y and Z form an equilateral triangle.
The resulting equilateral is known as the first Morley triangle, or simply the Morley triangle
(See Figure 3.1). The length of each side of the Morley triangle is equal to
XY = Y Z = XZ = 8r sin(α) sin(β) sin(γ)
where r is the circumradius of triangle ABC. The area of the Morley triangle can be found by
Area = 16
√
3r2
sin2
(α) sin2
(β) sin2
(γ)
The trilinear co-ordinates of each vertex of the Morley triangle are
X = 2 cos(β) : 2 cos(α) : 1
Y = 1 : 2 cos(γ) : 2 cos(β)
Z = 2 cos(γ) : 1 : 2 cos(α)
(Weisstein, ”Morley’s Theorem)
For the proof of theorem 3.1.1, we will need two lemmas, which we now state and prove before
proving the theorem.
29

31. Figure 3.1: Morley’s Theorem
Lemma 3.1.2. Let 0 < α < π. The function f(0, π − α) → R defined by
f(t) = sin(t)/ sin(t + α)
in the range 0 < t < π − α is injective.
Proof. We can prove Lemma 3.1.2 by using the contrapositive of the injection proof. Assume
f(x) = f(y), where f(x) = sin(x)/ sin(x + α) and f(y) = sin(y)/ sin(y + α)
sin(x)
sin(x + α)
=
sin(y)
sin(y + α)
Using the identity sin(A + B) = sin(A) cos(B) + cos(A) sin(B) on the denominator of each fraction
gives
sin(x)
sin(α) cos(x) + cos(α) sin(x)
=
sin(y)
sin(α) cos(y) + cos(α) sin(y)
Cross multiply the equation
sin(x) sin(α) cos(y) + sin(x) cos(α) sin(y) = sin(y) sin(α) cos(x) + sin(y) cos(α) sin(x)
We now need to start eliminating or simplifying some terms each side so that we are left with x = y.
Start by dividing the equation by sin(x) and then sin(y) (we’re assuming that sin(x), sin(y) = 0)
1
sin(y)
sin(α) cos(y) + cos(α) =
1
sin(x)
sin(α) cos(x) + cos(α)
Page: 30

32. We can cancel out cos(α) from both sides, and can also use the definition sin(A)/ cos(A) = cot(A)
to give
cot(y) sin(α) = cot(x) sin(α)
Now divide the equation by sin(α) to leave
cot(y) = cot(x)
As 0 < x < π − α and 0 < y < π − α, we can then conclude that
y = x
Thus proving the lemma to be true. QED
Lemma 3.1.3. For all real values of u, the following identity holds
sin(3u) = 4 sin(u) sin(π/3 + u) sin(π/3 − u) (3.1)
Proof. To prove Lemma 3.1.3, we need to show that the left hand side is equal to the right hand
side. For the left hand, we use De Moivre’s theorem, which is given by
(cos(x) + i sin(x))n
= cos(nx) + i sin(nx)
Substitute x = u and n = 3.
(cos(u) + i sin(u))3
= cos(3u) + i sin(3u)
Use binomial expansion on the left hand side
cos3
(u) + 3 cos2
(u)i sin(u) + 3 cos(u)i2
sin2
(u) + i3
sin3
(u) = cos(3u) + i sin(3u)
Separate real and imaginary parts (taking i2 = −1 and i3 = −i)
cos3
(u) − 3 cos(u) sin2
(u) + i(3 cos2
(u) sin(u) − sin3
(u)) = cos(3u) + i sin(3u)
Equate the imaginary parts
3 cos2
(u) sin(u) − sin3
(u) = sin(3u)
Use the identity cos2(u) + sin2
(u) = 1
3(1 − sin2
(u)) sin(u) − sin3
(u) = sin(3u)
Substitute s = sin(u)
3(1 − s2
)s − s3
= sin(3u)
Simplify
3s − 3s3
− s3
= sin(3u)
3s − 4s3
= sin(3u) (3.2)
Page: 31

33. For the right-hand side, we use the trigonometric identities
sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
sin(A − B) = sin(A) cos(B) − cos(A) sin(B)
Substituting A = u and B = π/3 into each identity
sin(π/3 + u) = sin(π/3) cos(u) + cos(π/3) sin(u)
sin(π/3 − u) = sin(π/3) cos(u) − cos(π/3) sin(u)
Take the product of these to give
(sin(π/3) cos(u) + cos(π/3) sin(u))(sin(π/3) cos(u) − cos(π/3) sin(u))
Expand the brackets
sin2
(π/3) cos2
(u)−sin(π/3) cos(π/3) sin(u) cos(u)+cos(π/3) sin(π/3) cos(u) sin(u)−cos2
(π/3) sin2
(u)
The second and third terms cancel out leaving
sin2
(π/3) cos2
(u) − cos2
(π/3) sin2
(u)
Substituting cos2(u) = 1 − sin2
(u), sin(u) = s, and calculating the known functions leaves
sin(π/3 + u) sin(π/3 − u) =
3
4
(1 − s2
) −
1
4
s2
(3.3)
Now we substitute equations (3.2) and (3.3) into (3.1)
3s − 4s3
= 4s(3
4(1 − s2
) − 1
4s2
)
Simplify and rearrange the equation to make both sides equal
3s − 4s3
= 4s(3
4 − 3
4s2
− 1
4s2
)
3s − 4s3
= 3 − 3s3
− s3
3s − 4s3
= 3s − 4s3
The left hand side equals the right hand side, thus proving the lemma is always true. QED
We will now prove Theorem 3.1.1, using the above lemmas. For convenience, the theorem has
been repeated here:
Theorem 3.1.1. Let ABC be a triangle with internal angles 3α, 3β and 3γ at vertices A, B and C
respectively. Let X be the intersection of the trisectors from A and B adjacent to side AB. Define
Y and Z similarly. Points X, Y and Z form an equilateral triangle.
Page: 32

34. Figure 3.2: Morley’s Theorem (with θ and µ)
Proof. Let a = BC, b = AC and c = AB and let θ = ∠AXZ and µ = ∠BXY , as in Figure 3.2. It
will be enough to show that
θ =
π
3
+ β (3.4)
and
µ =
π
3
+ γ (3.5)
because if this holds, then
∠Y XZ = 2π − (θ + µ + ∠AXB)
= 2π − (β + π/3 + α + π/3 + π − α − β)
= 2π − (2π/3 + π)
= π/3
We can do the same to show that ∠XZY and ∠XY Z are also equal to π/3, thus proving that XY Z
is an equilateral triangle. We can now prove equation (3.5). If we prove that equation (3.5) is true,
it will follow that equation (3.4) is also true, which will conclude the proof.
It will be sufficient to prove that
f(θ) = f(β + π/3) (3.6)
because f is injective from Lemma 3.1.2. From the sine rule we have
AZ
sin(θ)
=
AX
sin(θ + α)
Page: 33

35. because sin(∠XZY ) = sin(π − θ − α) = sin(θ + α). We can rearrange this to give
AZ
AX
=
sin(θ)
sin(θ + α)
= f(θ)
To complete the proof we must show that the ratio AZ/AX is also equal to f(π/3 + β). Because
3α + 3β + 3γ = π, we know that π − (α + γ) = 2π/3 − β. By using the sine rule again we have
AZ
sin(γ)
=
b
sin(π − α − γ)
=
b
sin(2π/3 + β)
=
b
sin(π/3 − β)
(3.7)
Using the same process with AX we get
AX
sin(β)
=
c
sin(π/3 − γ)
(3.8)
Dividing equation (3.7) by equation (3.8) we get
f(θ) =
AZ
AX
=
b sin(γ) sin(π/3 − β)
c sin(β) sin(π/3 − γ)
(3.9)
Setting t = π/3 + β, we have sin(t + α) = sin(π/3 + γ), therefore
f(π/3 + β) =
sin(π/3 + β)
sin(π/3 + γ)
(3.10)
and by using the sine rule we get
b sin(3γ) = c sin(3β) (3.11)
Using the identity in Lemma 3.1.3 on equation (3.11), with u = β on the left-hand side and
u = γ on the right-hand side, we calculate
b sin(π/3 + γ) sin(π/3 − γ) sin(γ) = c sin(π/3 + β) sin(π/3 − β) sin(β)
We can rearrange this to give
sin(π/3 + β)
sin(π/3 + γ)
=
b sin(γ) sin(π/3 − γ)
c sin(β) sin(π/3 − β)
The left-hand side is equal to equation (3.10) and the right hand side is equal to equation (3.9),
concluding that
f(π/3 + β) = f(θ)
As required from equation (3.6). QED
(Barbara, 1997: p447-450)
More Morley Triangles
Theorem 3.1.1 trisects the internal angles of triangle ABC, however, the theorem holds true when
trisecting the external angles of ABC too (pink triangle in Figure 3.3). The sides of this new Morley
triangle are parallel to the sides of the first Morley triangle. Furthermore if we take the intersections
of the pairs of remaining trisectors with the extended sides of this new Morley triangle, we get three
more equilateral triangles (red triangles in Figure 3.3). (Jost and Maor, 2014: p160-163)
Page: 34

36. Figure 3.3: More Morley Triangles
3.2 Napoleon’s Theorem
Let ABC be a triangle. Construct equilateral triangles (BCD, ACE, ABF) on each side of ABC,
all facing outwards. The centroids of these triangles, G, H and I have a special property (See Figure
3.4).
Theorem 3.2.1. If an equilateral triangle is constructed on each side of any triangle ABC, all
outwardly or all inwardly, then the centroids of these constructed triangles form themselves an
equilateral triangle.
This theorem is named after the French emperor Napoleon Bonaparte (1769-1821). He was an
amateur mathematician with an interest in geometry. However, there is much doubt as to whether
it was actually Napoleon who discovered this theorem. The earliest known reference to Napoleon’s
theorem is an entry from the 1825 edition of the Ladies’ Dairy written by William Rutherford
(1798-1871). He did not prove the theorem, but others provided a proof in the following year’s
edition. Neither edition mentioned Napoleon, he wasn’t attributed to the theorem until 1911 in
Elementi di Geometria. (Coxeter and Greitzer, 1967: p60-65)
Proof. Because angle IAC = GAB = 30◦, we can use the law of cosines to show that
GI2
= AI2
+ AG2
− 2 · AI · AG · cos(A + 60◦
) (3.12)
The centroid lies on the median 2/3 of the distance from vertex to the side of the triangle. Therefore
we get
AG = (2/3)
√
3/2 · AB = AB/
√
3
Page: 35

37. AI = (2/3)
√
3/2 · AC = AC/
√
3
and equation (3.11) becomes
3 · GI2
= AC2
+ AB2
− 2 · AC · AB · cos(A + 60◦
) (3.13)
By using the identity cos(A + B) = cos(A) cos(B) sin(A) sin(B), and that cos(60◦) = 1/2 and
sin(60◦) =
√
3/2, we get
3 · GI2
= AC2
+ AB2
− AC · AB · cos(A) +
√
3 · AC · AB · sin(A)
Using the law of cosines on ABC
BC2
= AC2
+ AB2
− 2 · AC · AB · cos(A)
and the law of sines
Area(ABC) = (1/2)AC · AB · sin(A)
we can write equation (3.12) as
3 · GI2
= (1/2)(BC2
+ AC2
+ AB2
) + 2
√
3 · Area(ABC)
Since this is symmetrical in BC, AC and AB, it follows that the triangle connecting the centroids
is equilateral. QED
(Bogomolny, n.d.)
Figure 3.4: Napoleon’s Theorem
Page: 36

38. Chapter 4
The Nine-Point Circle and The Euler
Line
4.1 The Nine-Point Circle
In 1821, Jean-Victor Poncelet (1788-1867) and Charles-Julien Brianchon (1783-1864), discovered
and proved a special property relating to nine particular points of any triangle.
Theorem 4.1.1 (Nine-Point Circle Theorem). If ABC is a triangle, then the midpoints of each
sides, the feet of the three altitudes, and the midpoints between each vertex and the orthocentre of
ABC all lie on a single circle.
(Coxeter, 1961: p18-20; Coxeter and Greitzer, 1967: p20-22; Venema, 2013: p57-59)
Definition 4.1.2 (Nine-Point Circle). The circle in Theorem 4.1.1 is known as the nine-point
circle.
(See Figure 4.1)
We will now prove that the nine points from Theorem 4.1.1 lie on a single circle.
Proof. Let ABC be a triangle. Let D, E and F be the midpoints of BC, AC and AB respectively,
J, K and L be the feet of the altitudes on BC, AC and AB respectively, H the orthocentre of
ABC, and X, Y and Z be the midpoints of AH, BH and CH respectively. We will now prove that
D, E, F, J, K, L, X, Y and Z lie on a single circle.
In ABC, since E and F are the midpoints of AC and AB, EF must be parallel to BC. In
BCH, Y and Z are the midpoints of BH and HC, therefore Y Z and BC must be parallel. It
follows that EF and Y Z are also parallel.
In BAH, since Y and F are midpoints of BH and BA, Y F is parallel to HA. In CAH, E and
Z are midpoints of AC and HC, so EZ is parallel to HA. Therefore it follows that Y F and EZ
are parallel. Since Y F is parallel to HA, and HA lies on AJ, Y F is also parallel to AJ.
37

39. As AJ is perpendicular to BC (since AJ is the altitude from A to BC), and BC and EF are
parallel, then AJ is also perpendicular to EF. Since Y F is parallel to AJ, Y F is perpendicular to
EF. Similarly, EZ is also perpendicular to EF. We now have a rectangle EFY Z, which can be
inscribed in a circle.
We can use a similar process as above to get the rectangle DFXZ, which can also be inscribed
in a circle. Because the two rectangles share opposite points F and Z, the two circles share the
same diameter FZ, and therefore coincide. From this D, E, F, X, Y and Z all lie on a common
circle.
Since AJ is an altitude of ABC from A to BC and XJ lies on AJ, ∠XJD is a right angle. As
XD is a diagonal of DFXZ, it is a diameter of the circle, so it follows that J must also lie on the
circle. Similarly for K and L.
Therefore all nine points D, E, F, J, K, L, X, Y and Z lie on the circle. This circle is the
nine-point circle.
QED
(Umberger, n.d.)
Definition 4.1.3 (Nine-Point Centre). The centre of the nine-point circle is called the nine-point
centre of ABC.
The nine-point centre is the midpoint of the line segment connecting the orthocentre and cir-
cumcentre of ABC. It is also considered a triangle centre, and appears in Kimberling’s Encylopedia
of Triangle Centres as X5, and is often labelled as N. The radius of the nine-point circle is equal
to half the radius of the circumcircle. (Coxeter, 1961: p18-20; Coxeter and Greitzer, 1967: p20-22)
The trilinear co-ordinates of the nine-point centre are
cos(B − C) : cos(C − A) : cos(A − B)
or, by using the sides of ABC,
bc(a2
(b2
+ c2
) − (b2
− c2
)2
) : ca(b2
(c2
+ a2
) − (c2
− a2
)2
) : ab(a2
(a2
+ b2
) − (a2
− b2
)2
)
Feuerbach’s Theorem
In 1822, Karl Wilhelm Feuerbach (1800-1834) published a theorem relating to the nine-point circle
and the four equicircles. These equicircles are the incircle and the three excircles.
Theorem 4.1.4 (Feuerbach’s Theorem). The Nine-Point Circle of triangle ABC is tangent to the
incircle and the three excircles of ABC.
The point where the nine-point circle and the incircle are tangent to each other is known as the
Feuerbach Point. This point is a triangle centre and is listed in the encylopedia as X11, and often
labelled F. The three points where the nine-point circle and the excircles are tangent to each other
forms the Feuerbach Triangle (See Figure 4.2). (Venema, 2013: p60-61; Weisstein, ”Feuerbach’s
Theorem”)
Page: 38

40. Figure 4.1: Nine-Point Circle and Nine-Point Centre
4.2 The Euler Line
Leonhard Euler (1707-1783) was a Swiss mathematician. Born in Switzerland, he spent the majority
of his life in Berlin, Prussia (now Germany) and St Petersburg, Russia. Despite many tragedies
happening throughout his later life, including the death of his wife, the loss of his home and library
through fire and becoming completely blind, his creative and academic abilities continued until his
death. By this time, he completed many works and discovered many formulas and theorems covering
nearly every area of mathematics, as well as science. In 1765, Euler discovered a relationship between
three particular triangle centres. (Jost and Maor, 2014: p123-125)
Theorem 4.2.1 (Euler Line Theorem). For any non-equilateral triangle, the centroid (G), the
circumcentre (O) and the orthocentre (H) are collinear. G lies between H and O and HG = 2GO.
Definition 4.2.2 (Euler Line). The line passing through G, H and O of any non-equilateral triangle
is called the Euler Line.
(See Figure 4.3)
However, since Euler discovered this line, it has been shown that other centres lie on the Euler
Line as well, most notably the Nine-Point Centre and the Exeter Point. Other points on the Euler
Line include the Schiffler Point, the de Longchamps Point and many more. (Coxeter, 1961: p17-18;
Coxeter and Greitzer, 1967: p18-20; Venema, 2013: p27-29)
Page: 39

41. Figure 4.2: Feuerbach’s Theorem
Figure 4.3: The Euler Line
Page: 40

42. Any point with trilinear co-ordinates α : β : γ lies on the Euler line if
α β γ
cos(A) cos(B) cos(C)
cos(B) cos(C) cos(C) cos(A) cos(A) cos(B)
= 0
which can also be written as
α cos(A)(cos2
(B) − cos2
(C)) + β cos(B)(cos2
(C) − cos2
(A)) + γ cos(C)(cos2
(A) − cos2
(B)) = 0
similarly
α sin(2A) sin(B − C) + β sin(2B) sin(C − A) + γ sin(2C) sin(A − B) = 0
(Weisstein, ”Euler Line”)
Right-Angled Triangle
If the triangle is right-angled, the Euler Line passes through the right-angled vertex and the midpoint
of the hypotenuse. This is because the orthocentre of the triangle coincides with the right-angled
vertex and the circumcentre coincides with the midpoint. Therefore, the Euler line also coincides
with the median through the right-angled vertex.
Isoceles Triangle
If the triangle is isoceles, the Euler line coincides with the single axis of symmetry of the triangle.
Also, the Euler passes through the Incentre (I) of the triangle, which it doesn’t otherwise.
Equilateral Triangle
If the triangle is equilateral, all the triangle centres coincide. The Euler Line is therefore not
defined in this case, which is why Theorem 4.2.1 and Definition 4.2.2 states that the triangle must
be non-equilateral.
Schiffler’s Theorem
Theorem 4.2.3 (Schiffler’s Theorem). For any triangle ABC with incentre I, the Euler Lines of
the four triangles BCI, ACI, ABI and the original triangle ABC are concurrent at S.
Definition 4.2.4 (Schiffler Point). The point of concurrency in Theorem 4.2.3 is called the Schiffler
Point.
The Schiffler Point, S, is an entry in Kimberling’s Encylopedia of Triangle Centres as X21.
The trilinear co-ordinates of the Schiffler Point are
=
1
cos(B) + cos(C)
:
1
cos(C) + cos(A)
:
1
cos(A) + cos(B)
Page: 41

43. =
b + c − a
b + c
:
c + a − b
c + a
:
a + b − c
a + b
(Emelyanov and Emelyanova, 2003)
Figure 4.4: Schiffler’s Theorem
Page: 42

44. Conclusion
In this report, I have discussed many theorems relating to triangle geometry. I have talked about
the centre of a triangle, and the fact that there are multiple centres for a single triangle, including
the incentre, centroid, orthocentre and circumcentre. I have explained the processes to find each
centre along with a justification of its existence through a proof, and have also provided an image
for each. I also stated and proved Ceva’s Theorem and used it to prove the concurrency of certain
sets of three lines that determine some of the triangle centres.
I then moved on to focus on proving three sets of special points are collinear in four different
theorems, including Menelaus’ Theorem, Desargues’ Theorem and Pappus’s Hexagon Theorem. I
then went on to investigate certain equilateral triangles associated with an arbitrary triangle in
Morley’s Theorem and Napoleon’s Theorem. Then finally, I discussed the nine-point circle and the
Euler line to round up my report.
Before I started working on this project, the last time I learnt any Euclidean geometry was in
school at GCSE level. So it has been a long time since I have learnt anything new in this area.
Whilst working on this project, I learnt a lot of new information about triangle geometry that I
never would have thought of before. For example, I never considered before where the centre of the
triangle might be or whether there was more than one candidate for it. I found undertaking this
project both challenging, instructive and interesting.
To complete this project I used LaTeX to complete it, and GeoGebra to create the figures. I
have had experience of word-processing in LaTeX for the past few years, using it to write up nearly
all of my coursework at university. GeoGebra was new to me and turned out to be a relatively easy
program to learn. It proved very useful for drawing detailed and clear diagrams for this report.
There were, however, many more theorems relating to triangle geometry that I did not include
in this report. They include plenty more triangle centres, for instance the Fermat point, Gergonne
point and Torricelli point. There are also the medial and orthic triangles, as well as Miquel’s theorem
and the Miquel Points. This report is in no way a full account of everything related to triangle
geometry, and only scratches the surface of a well-developed and intriguing area of mathematics.
43

45. References
.
Barbara, R. (1997). The Mathematical Gazette, Volume 81. London: Bell and Sons
Bogomolny, A. (n.d.). Napoleon’s Theorem, Two Simple Proofs. [on-line] Available at http:
//www.cut-the-knot.org/proofs/napoleon.shtml [Accessed 26 January 2015]
Coxeter H.S.M. (1961). Introduction to Geometry. New York: John Wiley & Sons, Inc
Coxeter, H.S.M. and Greitzer, S.L. (1967). Geometry Revisited. Washington DC: Mathematical
Association of America
Emelyanov, L. and Emelyanova, T. (2003). A Note on the Schiffler Point. [PDF] Available at
http://forumgeom.fau.edu/FG2003volume3/FG200312.pdf [Accessed 1 April 2015]
Encyclopaedia Brittanica on-line. (2015). Menelaus of Alexandria. [on-line] Available at
http://www.britannica.com/EBchecked/topic/374905/Menelaus-of-Alexandria [Accessed 25
January 2015]
Faucette, W.M. (2007). Ceva’s Theorem and Its Applications. [PDF] Available at: http:
//www.westga.edu/~faucette/research/CevaApps.pdf [Accessed 23 November 2014]
Field, J.V. (1995). Girard Desargues. [on-line] Available at http://www-history.mcs.st-andrews.
ac.uk/Biographies/Desargues.html [Accessed 25 January 2015]
Floor (1998). The Napoleon Point and More. [on-line] Available at http://mathforum.org/
library/drmath/view/55042.html [Accessed 8 March 2015]
Jackson, F. and Weisstein, E.W. (n.d). ”Simson Line.” From MathWorld. Available at http:
//mathworld.wolfram.com/Simson-line.html [Accessed 20 January 2015]
Jost, E. and Maor, E. (2014). Beautiful Geometry. Woodstock: Princeton University Press
Kimberling, C. (c 1994). Clark Kimberling’s Encyclopedia of Triangle Centers. [on-line] (Up-
dated 2015) Available at: http://faculty.evansville.edu/ck6/encyclopedia/ETC.html [Ac-
cessed 1 March 2015]
Kimberling, C. (1994) Central Points and Central Lines in the Plane of a Triangle. Mathemat-
ics Magazine, Vol. 67, No. 3. [on-line] Available at: http://www.jstor.org/stable/2690608
[Accessed 21 March 2015]
44

46. O’Connor, J.J. and Robertson, E.F. (2012). Giovanni Benedetto Ceva. [on-line] Available at
http://www-history.mcs.st-andrews.ac.uk/Biographies/Ceva_Giovanni.html [Accessed 12 April
2015]
Umberger, S. (n.d). Proof of the Nine-Point Circle. [on-line] Available at http://jwilson.coe.
uga.edu/emt668/EMAT6680.2000/Umberger/EMAT6680smu/Assign4smu/nineptproof.html [Accessed
8 April 2015]
Venema, G.A. (2013). Exploring Advanced Euclidean Geometry With Geogebra. Washington
DC: Mathematical Association of America
Weisstein, E.W. (n.d). ”Centroid.” From MathWorld. Available at http://mathworld.wolfram.
com/Centroid.html [Accessed 29 November 2014]
Weisstein, E.W. (n.d). ”Circumcenter.” From MathWorld. Available at http://mathworld.
wolfram.com/Circumcenter.html [Accessed 29 November 2014]
Weisstein, E.W. (n.d). ”Euler Line.” From MathWorld. Available at http://mathworld.
wolfram.com/EulerLine.html [Accessed 11 March 2015]
Weisstein, E.W. (n.d). ”Excircles.” From MathWorld. Available at http://mathworld.wolfram.
com/Excircles.html [Accessed 8 April 2015]
Weisstein, E.W. (n.d). ”Exeter Point.” From MathWorld. Available at http://mathworld.
wolfram.com/ExeterPoint.html [Accessed 9 March 2015]
Weisstein, E.W. (n.d). ”Feuerbach’s Theorem.” From MathWorld. Available at http://mathworld.
wolfram.com/FeuerbachsTheorem.html [Accessed 10 March 2015]
Weisstein, E.W. (n.d). ”Incenter.” From MathWorld. Available at http://mathworld.wolfram.
com/Incenter.html [Accessed 28 November 29 November 2015]
Weisstein, E.W. (n.d). ”Kimberling Center.” From MathWorld. Available at http://mathworld.
wolfram.com/KimberlingCenter.html [Accessed 1 March 2015]
Weisstein, E.W. (n.d). ”Morley’s Theorem.” From MathWorld. Available at http://mathworld.
wolfram.com/MorleysTheorem.html [Accessed 7 March 2015]
Weisstein, E.W. (n.d). ”Napoleon Points.” From MathWorld. Available at http://mathworld.
wolfram.com/NapoleonPoints.html [Accessed 30 November 2014]
Weisstein, E.W. (n.d). ”Napoleon’s Theorem.” From MathWorld. Available at http://mathworld.
wolfram.com/NapoleonsTheorem.html [Accessed 27 January 2015]
Weisstein, E.W. (n.d). ”Orthocenter.” From MathWorld. Available at http://mathworld.
wolfram.com/Orthocenter.html [Accessed 28 November 2014]
Wilson, J (n.d). Proof of Ceva’s Theorem and its Converse. [on-line] Available at http://
jwilson.coe.uga.edu/emt668/EMAT6680.2001/Mealor/EMAT6690/Ceva’s%20theorem/cevaproof.
html [Accessed 25 October 2014]
Page: 45

47. Appendices
Approximately 9,460 words
46

48. Page: 47

49. Diary
.
Meeting 2nd October 2014: This was my first meeting with supervisor, Trevor Hawkes. We
introduced ourselves to each other, including what modules I am studying on my course. Trevor
started explaining some basic information about the project, giving some examples of interesting
facts about triangles.
2nd - 9th October: Was tasked to research at least three interesting facts about triangles,
including theorems, proofs, etc, with the aim of finding something that Trevor did not know about.
Facts found included medians, Apollonius Theorem, Napoleon’s Theorem and Centres (including
Euler Line).
Meeting 9th October: I showed my fact sheet to Trevor. The project was then discussed in
more detail, with some background information on Ceva’s Theorem.
9th - 16th October: Began research on Ceva’s Theorem. This mainly involved researching
and writing up a proof of the theorem in my own words, as if writing for the report. This was for
practice in researching and in writing with good structure and clarity for report.
Meeting 16th October: Trevor read through my write-up of Ceva’s Theorem proof, and
suggested improvements that can be made.
16th - 21st October: Improve the write-up using Trevor’s suggestions.
Meeting 21st October: Trevor looked over the improvements and discussed further improve-
ments that can be implemented to finalise the section on Ceva’a Theorem. He also gave an intro-
duction to Morley’s Theorem.
21st October - 11th November (No meetings due to illness): Finished off the proof of Ceva’s
Theorem. Started researching Morley’s Theorem, using notes provided by Trevor as information
for proofs. Started writing up a proof of the theorem in my own words. Trevor suggested a possible
structure for the report, providing ideas of various information to research and include in the report.
Meeting 11th November: Trevor read through my write-up of the Morley’s Theorem proofs
and suggested improvements.
11th - 20th November: Made improvements to Morley’s Theorem proofs, and also proved of
a Lemma involved in the proof by calculations, using hints provided by Trevor. Also used Ceva’s
Theorem to prove certain line triples are concurrent, using a mixture of research and my own
knowledge.
Meeting 20th November: Trevor looked through work from over the week. He then suggested
that I start writing the report proper. Provided certain topics to research and include as part of
chapter 1 of the report.
20th - 27th November: Started writing chapter 1, and also including work done over the
term so far.
Meeting 27th November: Looked through current work of chapter 1 and suggested improve-
Page: 48

50. ments
27th November - 4th December: Made improvements to work so far and continued with
chapter 1. Created a structure in the report, including chapter headers, a contents page and a title
page. Made a start with the Literature Review.
Meeting 4th December: Quick meeting explaining current progress
4th - 10th December: Prepared for End-Of-Term Report.
Meeting 10th December: Handed in End-Of-Term Report.
10th December 2014 - 9th January 2015 (Christmas break): Started chapter 2. Wrote
proofs for three theorems.
Meeting 9th January: Discussed work done over Christmas break. Suggested further im-
provements that can be made to Morley’s Theorem section.
9th - 16th January: Made improvements, and wrote a proof of a lemma. Changed layout for
section
Meeting 16th January: Suggested further improvements.
16th - 23rd January: Made improvements. Completed section for The Simson Line
Meeting 23rd January: Read through Chapter 2 and suggested improvements. Talk about
layout and numbering of theorems and definitions. Suggested mentioning bits of maths history in
each section.
23rd - 30th January: Added more detail to Chapter 2, including information about the
founders of each theorem. I also changed Menelaus’ Theorem to take into account the direction of
the lines. Started Napoleon’s Theorem and drew an image in GeoGebra.
Meeting 30th January: Suggested some improvements to be made regarding the layout of
the report, including making page numbers appear on each page. Also recommended a webpage
about the Fermat Point to read through
30th - 6th February: Improved the layout of the report as per supervisor’s recommendations.
Read through webpage on Fermat Point and highlighted important information.
Meeting 6th February: Suggested further improvements to the layout of the report.
6th February - 5th March (No meetings due to supervisor’s annual leave and illness): Added
automatic numbering to the theorems, lemmas and definitions. Also reformatted the proofs, to make
it more in-line with standard report styles. Started on Chapter 4 with the Nine-Point Circle. Also
added a new section to Chapter 1 regarding the Encyclopaedia of Triangle Centres.
Meeting 6th March: Catch up on work completed over past month
6th - 20th March (No meeting due to Poster Presentation): Researched the Euler Line, as
well as little things for rest of report
Meeting 20th March: Found a proof for Napoleon Points
20th - 27th March: Wrote a proof for generalisation of Napoleon Points and included trilinear
Page: 49

51. points throughout report.
Meeting 27th March: Looked through Chapter 2. Supervisor suggested improvements to be
made on explanation and proofs for Menelaus’ and Desargues’ Theorems.
27th - 2nd April: Improved Menelaus’ and Desargues’ Theorem sections to improve explana-
tion. Started working through all diagrams to ensure all are done and consistent in design. Started
a list of references.
2nd - 10th April: Worked through the list of references and included what topics come from
which source. Fleshed out Chapter 3 and 4.
Meeting 10th April: Looked through Chapter 4. Found more information for the Schiffler
Point.
10th - 17th April: Finished main body and added images. Added references to bibliography
and citations in report. Started writing literature review.
Meeting 17th April: Read through literature review and suggested improvements.
17th - 24th April: Completed literature review, abstract, introduction and conclusion, with
support from supervisor. Sent report to family for proof reading. Finished report ready for sub-
mission on Friday 24th April.
Page: 50

52. Certificate of Ethical Approval
Student:
Christopher Thomson
Project Title:
The Secret Life of Triangles
This is to certify that the above named student has completed the Coventry
University Ethical Approval process and their project has been confirmed and
approved as No Risk
Date of approval:
16 October 2014
Project Reference Number:
P27558