Anca Mustaţǎ
Euclidean
Geometry
Contents
1. Euclid’s geometry as a theory, 3
2. Basic objects and terms, 3
2.1. Angles, 4
2.2. The circle, 5
2.3. The polygon, 5
2.4. The triangle, 5
2.5. Parallel lines, 5
3. Axioms of Euclidean geometry, 5
3.1. Angle measures, 6
4. Angles around a few lines, 6
4.1. Angles around two lines, 6
4.2. Angles around three lines when two are parallel, 7
5. Polygons and tiling, 9
5.1. Sum of angles around a triangle, 9
5.2. Angles around a polygon, 9
5.3. Regular polygons, 10
5.4. Tessellations, 10
6. Congruent triangles, 10
7. Triangles in parallelograms, 13
7.1. Special parallelograms, 14
8. The Pythagorean theorem, 15
9. Special lines in a triangle, 17
9.1. Angle bisector, 17
9.2. Constructing angle bisectors, 17
9.3. The perpendicular bisector, 19
9.4. Altitudes, 20
9.5. Median, 24
10. Circles, 27
10.1. Secants and tangents, 27
10.2. Arcs and angles, 29
11. Similar triangles, 32
12. The nine-point circle, 34
13. Menelaus and Ceva theorems, 35
References, 36
Index, 37
Warning: please read this text with a pencil at hand, as you will need to draw
your own pictures to illustrate some statements.
2
3
1. Euclid’s geometry as a theory
God is always doing geometry
— Plato, according to Plutarch [6]
These words suggest the reverence with which this branch of mathematics was regarded
by thinkers in the ancient world. They saw geometry as managing to extract proportions,
order and symmetry from the seemingly chaotic nature, thus making its beauty accessible
to the reasoning mind. For us as well, geometry is a bridge from visual representations of
the world to abstract logical thinking. This makes it a wonderful education tool. Indeed,
since our perception of the world is embedded in sensorial experiences, what better way to
develop a solid basis for our abstract thinking than to combine our visual intuition with
logical deductions?
It is for these reasons that an ancient geometry text has been referred to as the most
famous and influential textbook ever written. The Elements is a collection of thirteen
mathematical books attributed to Euclid, who taught at Alexandria in Egypt and lived
from about 325 BC to 265 BC. This is the earliest known historical example of a mathematical theory based on the axiomatic and logical deduction method.
A mathematical theory consists of
• a set of basic objects described by definitions,
• a set of basic assumptions about these objects: the axioms, and
• a set of statements derived from the axioms by logical reasoning.
– the most important of these are theorems,
– while less important statements are propositions,
– and corollaries are direct consequences of some previous statement,
– and lemmas are helpful in proving further propositions or theorems.
Each theorem, proposition or lemma consists of a hypothesis (set of assumptions), which
is what we know, and a conclusion: what we have to prove. These should be followed
by a proof , meaning a chain of statements related by logical implications, which starts
from the hypothesis, combines it with the axioms and/or statements already proven, and
arrives at the conclusion.
2. Basic objects and terms
All human knowledge begins with intuitions, thence passes to concepts and
ends with ideas.
— Immanuel Kant
Kritik der reinen Vernunft, Elementarlehle
Euclid’s geometry assumes an intuitive grasp of basic objects like points, straight lines,
segments, and the plane. These could be considered as primitive concepts, in the sense
that they cannot be described in terms of simpler concepts.
A point is usually denoted by an upper case letter. A straight line is usually denoted
by a lower case letter. We will think of a line as a set of points. Usually denote a line by
any two points on it: d = AB.
4
A
B
d
b
b
Note: for a point C on a line AB, one point can be between the other two:
C
b
A
b
A
b
B
b
C
b
B
b
A
b
B
b
C
b
A point A on a line d divides the line into two half-lines, or rays. Two points A and B
on the line d determine the segment AB, made of all the points between A and B.
A
B
AB
b
b
If three or more lines intersect at a point, they are concurrent at that point. If three or
more points are on the same line, they are collinear. A line in a plane divides the plane
in two half-planes.
2.1. Angles. In a plane, consider two half-planes bounded by two lines concurrent at the
point O. The intersection of the two half-planes is an angle. The two lines are the legs,
and the point the vertex of the angle. A particular angle in a figure is denoted by three
\ of which the middle one, A, is at the vertex, and the other two along the
letters, as BAC,
legs. The angle is then read BAC.
b
A
b
B
α
b
C
The angle formed by joining two or more angles together is their sum. Thus the sum
of the two angles ABC, P QR is the angle formed by applying the side QP to the side
BC, so that the vertex Q falls on the vertex B, and the side QR on the opposite side of
BC from BA.
When the sum of two angles BAC, CAD is such that the legs BA, AD form one
straight line, they are supplements of each other.
When one line stands on another, and makes the adjacent angles at both sides of itself
equal, each of the angles is a right angle, and the line which stands on the other is a
perpendicular to it. Write b ⊥ c to mean that a line b is perpendicular to a line c.
Hence a right angle is equal to its supplement. An acute angle is one which is less than
a right angle.
An obtuse angle is one which is greater than a right angle.
5
The supplement of an acute angle is obtuse, and conversely, the supplement
of an obtuse angle is acute.
When the sum of two angles is a right angle,each is the complement of the other.
2.2. The circle. A circle is a plane figure formed by a curve, the circumference, which
contains at least one point, and so that there is some point of the plane, the centre of the
circle, so that all segments drawn from the centre to any point of the circumference are
congruent to one another. Moreover, a point belongs to the circumference of the circle just
when the segment joining the point to the centre is congruent to one, hence to any, of those
segments. A radius of a circle is any segment drawn from the centre to the circumference.
A diameter of a circle is any segment drawn through the centre and terminated both ways
by point of the circumference. Four or more points found on the same circle are concyclic.
2.3. The polygon. A figure bounded by three or more segments is a polygon. The
segments are the sides of the polygon.
A polygon of three sides is a triangle. A polygon of four sides is a quadrilateral. A polygon
which has five sides is a pentagon; one which has six sides is a hexagon, and so on.
2.4. The triangle. A triangle whose three sides are unequal is scalene; a triangle having
two sides equal is isosceles; and, having all its sides equal, is equilateral. A right-angled
triangle is one that has one of its angles a right angle. The side which subtends the right
angle is the hypotenuse. An obtuse-angled triangle is one that has one of its angles obtuse.
An acute-angled triangle is one that has its three angles acute. An exterior angle of a
triangle is one that is formed by any side and the continuation of another side. Hence
a triangle has six exterior angles; and also each exterior angle is the supplement of the
adjacent interior angle.
2.5. Parallel lines. Two straight lines in the plane are parallel if they don’t meet.
3. Axioms of Euclidean geometry
1) A unique straight line segment can be drawn joining any two distinct points.
2) Any straight line segment is contained in a unique straight line.
3) Given any straight line segment, a unique circle can be drawn having the segment
as radius and one endpoint as center.
4) All right angles are congruent.
5) Given a point not on a given line, there exists a unique line through that point
parallel to the given line.
In truth, Euclid’s axioms are not sufficient for formally deducing the theorems of Euclidean geometry, or even for defining notions like “equal things,” or for comparing angles
and segments. Apart from the axioms, Euclid also relied on other “common sense” intuitive notions like rigid motion, boundary, interior and exterior of a figure, and so on.
The notion of rigid motion is necessary when comparing geometric objects. A rigid
motion of a geometric figure in plane can be understood as cutting the figure out of a sheet
of paper representing the plane and placing it in a different place in the plane. In practice,
we don’t cut figures out in order to move them – we clone them (copy them exactly) by
means of markings on a ruler (for segments) or protractor (for angles). As a basic tenet
6
of Euclidean geometry, you can thus move any geometric figure found somewhere in the
plane to any other position in plane. Interestingly, Euclid put effort into proving this
tenet for rigid motions of segments, while taking it for granted in the case of angles. Rigid
motion by means of a ruler and protractor is so ingrained in our way of doing geometry
that we don’t even notice how the notion of measure (centimeters, meters, inches etc. for
segments, and degrees for angles) is in fact an indirect process resulting from being able
to compare and add objects by moving them in the correct places.
Two geometric figures X and Y are congruent if one can move the first figure and
superpose it exactly on top of the second figure, such that the points of the two figures
now coincide. In this case we write X ≡ Y . A segment AB is smaller than another one
CD if one can move the segment AB until A coincides with D and B is in between C and
D. Similarly, an angle AOB is smaller than CO′ D if one can move AOB such that O
falls over O′ , the line OA over O′ C ′ , and B is in the interior of the angle CO′ D.
We will need to assume that, for any line AB, and real number x ≥ 0, there is a unique
point of that line that lies at distance x from A in the direction toward B.
Even after gathering all these extra basic assumptions in a set of axioms, there would
be some work to be done. One would have to eliminate the superfluous assumptions, i.e.
those which can be considered as theorems or propositions based on the other axioms.
For example, we do not need to assume rigid motion for all figures—only for angles and
segments. On the other hand, Euclid proved that a segment can be moved to any other
position if we assume that two circles, each passing through the interior of the other, intersect. Another problem may appear if some of the axioms introduced actually contradict
other axioms. To prove that the axioms are not contradictory, one would have to construct
a model of the plane for which all the axioms hold true, using other known mathematical
objects like numbers, vector spaces, etc.
Towards the end of the 19th century, David Hilbert began an immense effort to construct a sound axiomatic basis for each area of mathematics. His lectures at the university
of Göttingen in 1898–1899, published under the title Foundations of Geometry, proposed a larger set of axioms substituting the traditional axioms of Euclid. Hilbert proved
that his axioms are independent and non-contradictory (relying on algebra and coordinates
to construct a model of the plane satisfying his axioms). Since then, the algebraic/analytic
approach to Euclidian geometry has become dominant. Time permitting, we will discuss
Hilbert’s approach towards the end of the course.
Independently and contemporaneously, a 19-year-old American student named Robert
Lee Moore published an equivalent set of axioms. Some of the axioms coincide, while
some of the axioms in Moore’s system are theorems in Hilbert’s and vice-versa.
3.1. Angle measures. Define 1◦ as the 90-th part of a right angle, i.e. the measure of
an angle α such that 90 copies of α add up to a right angle. This definition makes sense
due to Euclid’s Postulate (4). Euclid doesn’t tell us that such an angle α exists! But our
intuition about the continuous nature of the plane tells us that α exists.
Degrees are defined based on the notion of right angles (and the assumption that they
are all equal), so if you try to define a right angle as being 90◦ , your definitions would be
moving in circles. Similarly if you tried to define supplements as summing up to 180◦ .
4. Angles around a few lines
4.1. Angles around two lines.
Proposition 4.1 (Opposite angles). Consider a line AB, a point O on it in between A
and B, and two points C and D in plane, on each side of the line AB respectively. Then
\ = BOD.
\
the points C, D and O are collinear if and only if AOC
7
b
C
B
b
O
b
Ab
D
b
Proof. Here “If and only if” means that you have to prove two statements:
\ = BOD
\ are equal.
=⇒ If the points C, D and O are collinear then the angles AOC
\ = BOD
\ then C, D and O are collinear.
⇐= If AOC
\ and
Prove 4.1 by first noticing that supplements add up to 180◦ , and then that both AOC
\ are supplements of the same angle BOC.
\ Prove 4.1 by contradiction. Assuming C,
BOD
D and O are not collinear, extend the line CO on the other side of AB by OE and then
\ = BOE.
\ Argue (using one of the common notions) that in
use 4.1 to prove that BOD
this case the lines OE and OD should coincide.
4.2. Angles around three lines when two are parallel. Remember Euclid’s Postulate
5): Given a point not on a given line, there exists a unique line through that point parallel
to the given line. In general, if a straight line l intersects two other straight lines a and b,
the sum of the interior angles on the same side of l satisfies one of the following properties:
sum = 180◦
sum > 180◦
sum < 180◦
In the first case, we expect that the lines do not intersect. Two lines a and b in the plane
which do not intersect (no matter how far we extend them) are parallel, denoted a k b.
Theorem 4.2 (Parallel lines I). Let AB and CD be two distinct lines crossed by another
\
line at the points P and Q like in the figure below. Then AB k CD if and only if BP
Q+
◦
\
P
QD = 180 .
b
b
C
A
P
b
b
Q
b
b
B
D
Proof. The proof has two parts.
\
\
=⇒ We assume that BP
Q+P
QD = 180◦ and prove AB k CD. Proof by contradiction: Assume AB ∦ CD. Then AB meets CD at a point M , on one side of the
line P Q, for example on the same side as B and D.
8
b
N
b
A
b
P
b
B
b
M
b
b
C
D
b
Q
Then on the other side of the line P Q we can construct a point N such that
M QP ≡ N P Q. Indeed, it is sufficient to choose N on the line AB such that
|P N | = |QM |. Then
\
\
\
M
QP = 180◦ − M
PQ = N
P Q,
the last equality being due to the fact that M, B, P, A, N form a line. Thus by
\
\
placing the angle M
QP on top of the angle N
P Q so that Q is placed on top of
P and P on top of Q, then M QP can fit exactly on top of N QP which means
that the two triangles are congruent.
On the other hand, this implies that
\
\
\
\
N
QP = M
P Q = 180◦ − N
P Q = 180◦ − M
QP ,
so the points M, D, Q, C, P are collinear. Since M, B, P, A, N also form a line,
this would mean that the lines AB and CD are not distinct, which contradicts
the hypothesis. The contradiction is due to our assumption that AB ∦ CD. It
follows that AB k CD.
\
\
⇐= We assume AB k CD. We prove BP
Q+P
QD = 180◦ . Proof by contradiction:
\
\
We can construct a line P E such that BP
Q+P
QE = 180◦ , with D and E on
the same side of line P Q. Then by Part 4.2, it follows that AB k QE. As we
know AB k QD and Euclid’s 5th Postulate states that through the point Q there
should pass a unique line parallel to AB, it follows that Q, E, D are collinear and
\
\
\
\
hence P
QD = P
QE. Thus BP
Q+P
QD = 180◦ as required.
Theorem 4.3 (Parallel lines II). The following statements are equivalent:
• AB k CD
\
\
• BP
Q+P
QD = 180◦ : two interior consecutive angles add up to 180◦ .
\
\
• AP
Q+P
QC = 180◦ : two interior consecutive angles add up to 180◦ .
\
\
• P
QD = AP
Q: two alternate angles are equal.
\
\
• BP
Q=P
QC: two alternate angles are equal.
\
\
• EP
A=P
QC: two corresponding angles are equal.
\
\
• EP
B=P
QD: two corresponding angles are equal.
\
\
• QP
B=F
QD: two corresponding angles are equal.
\
\
• F
QC = QP
A: two corresponding angles are equal.
b
A
P
b
Q
C
b
b
b
F
E
B
b
b
D
b
9
Proof. Theorem 4.2 proves that the first three statements are equivalent to one another.
To prove that the second holds if and only if the fourth holds, note that both
\
\
\
\
AP
Q + QP
B = 180◦ and P
QD + QP
B = 180◦ .
All of the other equivalences, can be proven in a similar way.
5. Polygons and tiling
5.1. Sum of angles around a triangle.
Theorem 5.1. The sum of all the interior angles of a triangle is 180◦ .
b
D
b
B
A
b
b
E
b
C
Proof. Consider ABC. There exists a unique line DE passing through A such that DE k
BC, like in the diagram. Using the crossing lines AB and AC, the previous theorem
implies:
\ = BAD
\ and ACB
\ = CAE
\
ABC
(pairs of alternate angles). Then
\ + BAC
\ + ACB
\ = BAD
\ + BAC
\ + CAE,
\
ABC
\
= DAE,
= 180◦ .
5.2. Angles around a polygon.
Theorem 5.2. The sum of the angles around an n-sided polygon is 180◦ (n − 2).
Proof. Denote by S(n) the sum of all the sizes of all interior angles of a polygon with n
sides.
n angle
3
4
5
180◦
360◦
540◦
Indeed, S(3) = 180◦ is known; the interior of a quadrilateral can be split into 2 triangles,
and a pentagon into 3 triangles. We can thus prove
(5.1)
S(n) = (n − 2)180◦
by induction on n. For the induction step, we assume S(n) = (n − 2)180◦ and need to
prove S(n + 1) = (n − 1)180◦ . For this, we split the interior of a polygon of (n + 1) sides
into one of n sides and a triangle. Thus S(n + 1) = S(n) + 180◦ from our construction,
and S(n) = (n − 2)180◦ from our assumption. Putting these together we get S(n + 1) =
(n − 1)180◦ .
10
5.3. Regular polygons. A polygon is regular if all of its sides are equal and all of its
angles are equal. Theorem 5.2 yields:
Corollary 5.3. Any regular polygon with n sides has n angles all equal to 180◦ (n − 2)/n:
sides
angles
60◦
90◦
108◦
120◦
3
4
5
6
5.4. Tessellations. A tessellation or tiling of the plane is a collection of plane figures
that fills the plane with no overlaps and no gaps. The classic two-dimensional picture of
a beehive is a tiling made out of regular hexagons. This makes the beehive into a sturdy
construction, comfortable for the bees and suitable for their communal life. But why don’t
the bees construct their beehives out of pentagonal, or octagonal shapes? In mathematical
terms, we could pose this question as follows: for which integer numbers n ≥ 3 does there
exist a tiling of the plane by identical regular n-sided polygons?
Theorem 5.4. The only tilings by regular and identical polygons are by equilateral triangles, by squares, or by hexagons.
Proof. Assume that there exists a tiling by identical regular n-sided polygons, and look at
all the angles around a vertex A of the tiling. Their sum is 360◦ , and they are all equal to
each other. By corollary 5.3, each angle is 180◦ (n − 2)/n. Assume that there are k angles
around some vertex A of the tiling, for an unknown integer k:
(n − 2)
2n
k
180◦ = 360◦ , or, after simplifying, k =
.
n
n−2
2n
Since k = n−2
is a whole number, n−2 divides 2n. But n−2 also divides 2(n−2) = 2n−4,
and so n − 2 divides 4, i.e. n − 2 = 1, 2 or 4, so n = 3, 4 or 6.
Exercise. Find all pairs of integer numbers m > n ≥ 3 such that there exists a tiling of
the plane made only of regular polygons of n sides and m sides. For each pair, make a
sketch of a possible tiling.
+ l m−2
= 2 case by case. Note that in this case k, l ≥ 1.
Hint: Solve the equation k n−2
n
m
5
Start with the case k = 1, n = 3. Then l m−2
= 35 or, equivalently, m−2
= 3l
. When
m
m
l = 1, 3 or 4, the equation has no integer solution m. When l = 2, we get m = 12. When
l = 5 we get m = n = 3. If l > 5 then m < 3 which contradicts our basic assumption
about m.
Continue with the case k = 2, n = 3 etc.:
k
n
l
m
1
2
3
4
1
3
3
3
3
4
2
2
2
1
2
12
6
4
6
8
6. Congruent triangles
′
Two triangles ABC and A B ′ C ′ are congruent if one can be superposed exactly on the
other such that the point A coincides with A′ , the point B with B ′ and the point C with
C ′ . In this case we write
ABC ≡ A′ B ′ C ′ .
11
Corresponding parts of congruent triangles are congruent:
|AB| = |A′ B ′ |,
|AC| = |A′ C ′ |,
|BC| = |B ′ C ′ |,
 = Â′ ,
B̂ = B̂ ′ ,
Ĉ = Ĉ ′ .
Theorem 6.1 (Side-Angle-Side or SAS). If two triangles ABC and A′ B ′ C ′ satisfy
A
b
B
A′
b
b
B ′b
C
b
C′
b

|AB| = |A′ B ′ | 
then: ABC ≡ A′ B ′ C ′
 = Â′
′ ′ 
|AC| = |A C |
Proof. Since  = Â′ , we can superpose the two angles such that the rays AB and A′ B ′
coincide, and the rays AC and A′ C ′ coincide. Then on the ray AB, we measure a length
segment |AB| = |A′ B ′ |. This will place B ′ exactly in the same place as B. Similarly for
C and C ′ .
Theorem 6.2 (Angle-Side-Angle or ASA). If two triangles ABC and A′ B ′ C ′ satisfy
b
B
A
b
b
b
A′
C B′ b
b
C′


 = Â′
|AB| = |A′ B ′ |
then: ABC ≡ A′ B ′ C ′ .

′
B̂ = B̂
Proof. Since |AB| = |A′ B ′ |, we can superpose the two segments such that A and A′
coincide, and B and B ′ coincide. Then since  = Â′ , we can superpose the ray A′ C ′ on
AC. Since B̂ = B̂ ′ , we can superpose the ray B ′ C ′ on BC. Thus C, the intersection of
AC and BC, will coincide with C ′ , the intersection of A′ C ′ and B ′ C ′ will coincide with
.
As an application we have:
Theorem 6.3 (Isosceles triangle). In any triangle ABC, the ABC is isosceles with
|AB| = |AC| if and only if B̂ = Ĉ.
A
b
B
Proof.
b
b
C
12
=⇒ We assume ABC is isosceles with |AB| = |AC|. We prove B̂ = Ĉ by comparing
ABC with its “flip” ACB as follows:

|AB| = |AC| 
\
\ = ACB
SAS: ABC ≡ ACB
ABC

|BC| = |CB|
Thus also B̂ = Ĉ.
⇐= We assume B̂ = Ĉ. We then prove |AB| = |AC| by showing that ABC ≡ ACB
by (ASA).
Theorem 6.4 (Side-Side-Side or SSS). If two triangles ABC and A′ B ′ C ′ satisfy
|AB| = |A′ B ′ |
|AC| = |A′ C ′ |
|BC| = |B ′ C ′ |
)
then: ABC ≡ A′ B ′ C ′ .
Proof. With no assumptions about angles, we cannot complete the superposing argument
as before. We will then return to proof by contradiction. Place A′ B ′ C ′ on the same side
of the line AB as ABC, such that the segment A′ B ′ ] is superposed on AB], i.e. such that
B ′ coincides with B and A′ with A, but assume that C 6= C ′ . This can happen in several
cases:
a) If C ′ is in the interior of the triangle ABC.
b
b
A
b
C
C′
b
B
From the assumption SSS, the triangles ACC ′ and BCC ′ are isosceles with |AC| =
|AC ′ | and |BC| = |BC ′ |, which by the previous Theorem implies
′ C and BCC
′ C.
\′ = AC
\
\′ = BC
\
ACC
Adding up:
(6.1)
′ C + BC
′ C = 360◦ − AC
′ B.
\ = ACC
\′ + BCC
\′ = AC
\
\
\
ACB
′ B < 180◦ , so 360◦ − AC
′ B > 180◦ .
\ < 180◦ and on the other hand AC
\
\
But ACB
′
Contradiction with the equality 6.1! Thus C is not in the interior of the triangle
ABC.
b) If C is in the interior of the triangle ABC ′ , we repeat the same proof as in Case
1., after swapping C ′ and C.
′ C and BCC
′ C.
\′ = AC
\
\′ = BC
\
c) Like in Part a), ACC
13
b
A
C′
b
C
b
b
B
′ CB − C
′ CA while swapping the role of C and C ′ gives
\=C
\
\
So ACB
′ B = CC
′ B − CC
′ A,
\
\
\
AC
′ CB − C
′ CA,
\
\
=C
\
= −ACB.
′ B = −ACB
\
\ which is impossible, since both angles are positive.
So AC
7. Triangles in parallelograms
A quadrilateral whose opposite sides are parallel is a parallelogram.
Theorem 7.1 (Parallelogram). Let ABCD be a quadrilateral. The following statements
are equivalent:
a) ABCD is a parallelogram.
b) AB k CD and |AB| = |CD|.
c) The diagonals AC and BD intersect at their midpoint.
d) |AB| = |CD| and |BC| = |AD|.
B
b
b
b
A
b
C
E
b
D
Proof. The four statements a)-d) are equivalent if each of them implies any other among
them. Thus it seems that we would have to prove a total of 12 implications! (check that
14
12 is the correct number). However, we can considerably shorten our work by following:
a)
b)
d)
c)
The top arrow means that we will assume a) and prove that it implies b), and so on.
\ = DBC.
\
a) =⇒ b) We only need to prove |AB| = |CD|. Note that BC k AD so BDA

\ = CDB
\ 
ABD
|BD| = |DB|
ASA: BDA ≡ DBC
\ = DBC
\ 
BDA
So |AB| = |CD|.
b) =⇒ c) Let E denote the intersection point of the diagonals. Alternate angles in ABkCD
\ = CDB.
\ Alternate angles in ABkCD give BAC
\ = DCA.
\
give ABD

\ = ECD
\ 
EAB
|AB| = |CD|
ASA: ABE ≡ CDE
\ = CDE
\ 
ABE
so |AE| = |CE| and |BE| = |DE|.
\ = DEC.
\
c) =⇒ d) By opposite angles, BEA

|AE| = |CE| 
\
\
SAS: AEB ≡ CED
AEB = CED

|EB| = |ED|
so |AB| = |CD|. Similarly, we can prove CBE ≡ ADE and hence |CB| = |AD|.
d) =⇒ a)
|AB| = |CD|
|BC| = |DA|
|AC| = |CA|
)
SSS: ABC ≡ CDA
\ = DCA
\ and ACB
\ = CAD.
\ So AB k CD and BC k AD by the theorem
So BAC
on parallel lines.
7.1. Special parallelograms. A quadrilateral with four right angles is a rectangle. A
rectangle is a parallelogram by the Theorem on parallel lines (4.3).
Theorem 7.2 (Rectangle). Let ABCD be a quadrilateral. The following statements are
equivalent:
a) ABCD is a rectangle.
b) The diagonals AC and BD intersect at their midpoint and |AC| = |BD|.
Proof.
=⇒ By the theorem on parallelograms, |AB| = |CD|.

So |AC| = |BD|.
|BA| = |CD| 
\ = CDA
\
SAS: BAD ≡ CDA
BAD

|AD| = |DA|
15
⇐= We first note that ABCD is a parallelogram by the Theorem on parallelograms
(C), since the diagonals AC and BD intersect at their midpoint. By the theorem
on parallelograms, |AB| = |CD|.
|AB| = |DC|
|BD| = |CA|
|AD| = |DA|
)
SSS: ABD ≡ DCA
\ = CDA
\ and, since the sum of the two angles is 180◦ by the Theorem on
So BAD
\ = CDA
\ = 90◦ . Then again, by the Theorem
parallel lines, it follows that BAD
\ + ABC
\ = 180◦ and CDA
\ + DCB
\ = 180◦ , so ABC
\ =
on parallel lines, BAD
◦
\ = 90 .
DCB
A quadrilateral all of whose sides are equal is a rhombus. Any rhombus is a parallelogram by the theorem on parallelograms d).
Theorem 7.3 (Rhombus). Let ABCD be a quadrilateral. The following statements are
equivalent:
a) ABCD is a rhombus.
b) The diagonals AC and BD intersect at their midpoint and AC ⊥ BD.
Proof.
a) =⇒ b) Triangles BAC and DAC are isosceles, so the angles opposite the equal sides are
equal. By SAS, the triangles are congruent.
\ = BCA
\ = DAC
\ = DCA
\ = 1 BAD.
\
(7.1)
BAC
2
Similarly, ABD and CBD are isosceles with |AB| = |BC| = |CD| = |DA| so
\ = ADB
\ = CBD
\ = CDB
\ = 1 ADC.
\
(7.2)
ABD
2
On the other hand, AB k CD so
\ + ADC
\ = 180◦ .
BAD
This, together with equations (7.1) and (7.2), implies
\ + ABD
\ = 90◦ .
BAC
a) =⇒ b) Let E be the intersection point of AC and BD.

|AE| = |CE| 
\ = CEB
\
SAS: AEB ≡ CEB
AEB

|EB| = |EB|
So |AB| = |BC|. By the Theorem on parallelograms, all sides are equal.
8. The Pythagorean theorem
In this section we will employ the notion of area of a bounded plane figure. We assume
that:
a) Any bounded plane figure has an area, which is a nonnegative real number.
b) The area of a rectangle is the product of its length and height.
c) Two congruent figures have the same area.
d) If we glue together two plane figures along finitely many line segments, then the
area of the union is the sum of the areas.
16
Lemma 8.1. The area of a triangle ABC with  = 90◦ is
1
|AB|
2
· |AC|.
Proof. We complete the triangle ABC to a rectangle ABDC. By the Theorem on rectangles above, ABC ≡ DCB, and as such, each of the triangles has an area equal to half the
area of the rectangle.
Theorem 8.2 (The Pythagorean theorem). Let ABC be a triangle with  = 90◦ . Then
|AB|2 + |AC|2 = |BC|2 .
Proof. On the ray AB, extend the segment AB by a segment BM such that |BM | = |AC|.
N
b
b
D
P
b
C
b
E
b
b
A
b
B
b
M
On the ray AC, extend the segment AC by a segment CN such that |CN | = |AB|.
Construct the square M AN P . Take points D from N P and E from P M so that
|M B| = |AC| = |N D| = |P E| and |M E| = |AB| = |N C| = |P D|.
These relations, together with M̂ = Â = N̂ = P̂ = 90◦ , imply
M EB ≡ ABC ≡ N CD ≡ P DE.
So
\
\=N
\
\
M
EB = ABC
CD = P
DE,
\
\=N
\
\
M
BE = ACB
DC = P
ED,
|BE| = |CB| = |DC| = |ED|.
\ = 180◦ − ABC
\−M
\
\ − ACB
\ = 90◦ , and similarly
From here, CBE
BE = 180◦ − ABC
\ = BCD
\ = CDE
\ = DEB
\ = 90◦ .
EBC
Thus BCDE is a square, since all its angles are right angles and all its sides are equal.
We can now compare areas:
Area AN P M = Area BCDE + 4 Area ABC,
1
so (|AB| + |AC|)2 = |AM |2 = |BC|2 + 4 · |AB| · |AC|,
2
so |AB|2 + |AC|2 = |BC|2 .
Corollary 8.3 (The congruence case). Two right angle triangles ABC and A′ B ′ C ′ having
two pairs of sides of equal lengths |AB| = |A′ B ′ | and |AC| = |A′ C ′ | are congruent triangles.
Proof. Applying Pythagoras’ theorem to both triangles will result in |BC| = |B ′ C ′ | as
well so we can use SAS or SSS.
17
9. Special lines in a triangle
In this section we will use triangle congruences to study five types of important lines in
a triangle: angle bisectors, perpendicular bisectors, altitudes, medians, and midlines. We
will prove that the three angle bisectors of a triangle intersect at a unique point. Similarly
for the three perpendicular bisectors; for the three altitudes and for the three medians.
We will also study the defining properties of the points forming these lines.
\ is the line AD such that D is
9.1. Angle bisector. The angle bisector of an angle BAC
\ and
a point in the interior of the angle BAC
\ = DAC.
\
BAD
C
b
D
b
A
B
b
b
9.2. Constructing angle bisectors. Draw an arc of circle centered at A, and let the
intersection points with the rays AB and AC be E and F , respectively. Two arcs of circles
centered at E and F respectively, and of the same radius, will intersect at a point D.
F
b
b
b
A
D
b
b
b
E
\ Indeed,
Then AD is the angle bisector of BAC.
|AE| = |AF |
|ED| = |F D|
|AD| = |AD|
)
SSS: AED ≡ AF D
\ = DAF
\.
So EAD
There is an alternative way of characterizing the bisector, involving the distance from
a point to a line.
The distance from a point D to a line AB not containing the point is the length |DM |,
where DM ⊥ AB and M is a point of AB. Note that this is the same as the shortest path
from D to a point on AB, due to the Pythgoraean theorem. A geometric locus is a set of
points in the plane all of which satisfy a given property.
\ and a
Theorem 9.1 (The angle bisector as a geometric locus). Consider an angle BAC
point D in its interior. The following two statements are equivalent:
\
a) AD is the angle bisector of BAC.
b) The point D is equidistant from AB and AC.
18
In other words, the angle bisector is the geometric locus of all the points in the interior
of an angle equidistant from the sides of the angle.
B
b
M
b
D
b
A
b
b
N
C
b
Proof. Consider the points M of AB and N of AC such that DM ⊥ AB and DN ⊥ AC.
=⇒ By right angles,
\ = 90◦ − DAM
\ = 90◦ − DAN
\ = ADN
\
ADM
so

\
\
M
AD = N
AD 
|AD| = |AD|
ASA: ADM ≡ ADN
\ = ADN
\ 
ADM
So the distances |M D| and |N D| from D to AB and AC, respectively, are equal.
⇐= By the Pythagorean Theorem,
|AM | =
p
p
|AD|2 − |M D|2 =
|M D| = |N D|
|DA| = |DA|
|M A| = |N A|
)
|AD|2 − |N D|2 = |AN |.
SSS: M DA ≡ N DA
\
\ . Thus AD is the angle bisector of M
\
\
So M
AD = DAN
AN = BAC.
This universal property of a bisector is helpful in proving an important property of a
triangle:
Theorem 9.2 (The incentre of a triangle). All of the angle bisectors of the interior angles
in a triangle ABC intersect at a point I, the incentre of the triangle.
b
A
b
N
E
b
M
b
I
b
B
b
b
P
b
b
C
D
\ and ABC
\ respectively,
Proof. We first notice that the angle bisectors AD and BE of BAC
must intersect at a point. Indeed, if they were parallel, then by the theorem on parallel
\ and EBA
\ = 1 ABC,
\ and so we would
\ + EBA
\ = 180◦ . But DAB
\ = 1 BAC
lines, DAB
2
2
19
\ + ABC
\ = 360◦ , which is absurd: they are interior angles in ABC and so their
have BAC
sum should be 180◦ .
We will denote the point of intersection of AD and BE by I. It remains to show that
\ Let |IM |, |IN |, and |IP | be the distances from I to AB,
CI is the angle bisector of ACB.
AC and BC respectively, with M on AB, N on AC, P on BC.
\ by the universal property of angle bisectors we
Since AD is the angle bisector of BAC,
have |IM | = |IN |.
\ by the universal property of angle bisectors we
Since BE is the angle bisector of ABC,
have |IM | = |IP |
Thus |IN | = |IP | and so, by the universal property of angle bisectors, CI is the angle
\
bisector of ACB.
In the proof above, we’ve seen that |IM | = |IN | = |IP |, and so, there exists a circle
with centre at I and radius IM , the incircle of ABC. Because IM ⊥ AB, IN ⊥ AC
and IP ⊥ BC, the incircle intersects the sides of ABC at the points M, N, P only. The
incircle is tangent to the sides of ABC, i.e. touches each side once. Indeed, if the incircle
intersected AB at two points M and M ′ , then IM M ′ would be isosceles with the angles
at M and M ′ both equal to 90◦ , which is impossible.
9.3. The perpendicular bisector. The perpendicular bisector of a segment BC is the
line perpendicular on BC and passing through its midpoint. To construct the perpendicular bisector, draw two circles centered at B and C respectively, and of the same radius,
which intersect at two points S and T . Then the line ST is the perpendicular bisector
of BC. By construction, |BS| = |CS| = |BT | = |CT | and so BSCT is a rhombus. By
the theorem on rhombi, we know that ST ⊥ BC and that ST and BC intersect at their
midpoints.
There is an alternative way of characterizing the perpendicular bisector.
Theorem 9.3 (The perpendicular bisector as a geometric locus). Consider a segment
BC and a point S not on it. The following two statements are equivalent:
a) S lies on the perpendicular bisector of BC.
b) |SB| = |SC|.
In other words, the perpendicular bisector of a segment is the geometric locus of all
the points equidistant from the vertices of the segment.
Proof. Consider the point M of BC such that SM ⊥ BC.
a) =⇒ b) By assumption, M is the midpoint of BC and SM ⊥ BC, thus

|BM | = |CM | 
\
\
SAS: BM S ≡ CM S
BM
S = CM
S

|M S| = |M S|
So |SB| = |SC|.
\
\
a) ⇐= b) By
B = SM
C = 90◦ . By Pythagoraean Theorem, |BM | =
p construction, SMp
|SB|2 − |M S|2 =
|SC|2 − |M S|2 = |CM |. Thus SM is the perpendicular
bisector of BC.
This universal property of a perpendicular bisector is helpful in proving an important
property of a triangle:
Theorem 9.4 (The circumcentre of a triangle). All the perpendicular bisectors of the
sides in a triangle ABC intersect at a point O, the circumcentre of the triangle, because
there exists a circle, the circumcircle of the triangle, of centre O and containing the vertices
of the triangle ABC.
20
A
b
b
b
O
b
B
b
b
b
C
Proof. We first notice that the perpendicular bisectors of BC and AB must intersect at
a point, which we will call O. Indeed, if they were parallel, then by drawing a common
parallel to both through B and applying the theorem on parallel lines, we would get
\ = 180◦ , which is absurd.
ABC
It remains to show that O lies on the perpendicular bisector of AC. Since O lies on
the perpendicular bisector of BC, by the universal property of perpendicular bisectors we
have |OB| = |OC|.
Since O lies on the perpendicular bisector of BA, by the universal property of perpendicular bisectors we have |OB| = |OA|.
Thus |OA| = |OC| and so, by the universal property of perpendicular bisectors, O lies
on the perpendicular bisector of AC.
Since |OA| = |OB| = |OC|, the point O is the centre of a circle which contains all
vertices of the triangle ABC.
Example 9.5. The circumcentre of a right angled triangle is always the midpoint of the
hypothenuse.
Indeed, we can complete ABC with ∡A = 90◦ to a rectangle ABA′ C. From the
theorem of rectangles, we know that the diagonals AA′ and BC are equal and intersect
at their midpoint O. Thus O is equidistant from A, B, C, A′ .
9.4. Altitudes. The altitude from the vertex A of a triangle ABC is the line through A
perpendicular on BC.
Example 9.6. The lines AD, EF and HK are altitudes in the three distinct triangles
below:
b
B
b
E
b
F
A
b
b
D
b
C
H
b
b
G
b
K
b
I
b
J
Theorem 9.7 (Orthocentre). All of the altitudes of a triangle intersect at a point H, the
orthocenter of the triangle.
21
C′
B′
A
b
b
P
B
b
b
b
N
b
b
b
M
b
C
A′
Proof. Through each of the vertices of the triangle ABC we draw a parallel to the opposite
side. By intersecting these lines we get another triangle A′ B ′ C ′ , such that A is on B ′ C ′ ,
B is on A′ C ′ , and C is on A′ B ′ .
The proof is based on the observation that the altitudes in the triangle ABC are
perpendicular bisectors in triangle A′ B ′ C ′ . Since we have shown that the perpendicular
bisectors of a triangle are concurrent, it will follow that the altitudes in the triangle ABC
are concurrent at a point H.
Indeed, ABCB ′ and ACBC ′ are parallelograms, so by the theorem on parallelograms,
|AB ′ | = |BC| and |BC| = |AC ′ |, which imply |AB ′ | = |AC ′ |. Moreover, since BC k
B ′ C ′ , it follows that the altitude AM of the triangle ABC is also perpendicular to B ′ C ′ ,
and passing through its midpoint A. Thus AM is the perpendicular bisector of B ′ C ′ .
Similarly, the altitude BN is the perpendicular bisector of A′ C ′ and the altitude CP is
the perpendicular bisector of A′ B ′ .
As in the case of angle and perpendicular bisectors, it would be nice if we could express
an altitude as a geometric locus, i.e. if we could decide whether a point H is on the
altitude from A to BC based on measurements involving only H, A, B and C. Given
a protractor, we could simply check whether the angle between the lines AH and BC is
90◦ . Given only a ruler, we could make use of Pythagora’s theorem to check whether the
same angle is 90◦ . Since our measurement should involve only segments made by H, A,
B and C, we will come up with a slightly complicated condition obtained by applying
Pythagoras’ theorem a number of times and cancelling some irrelevant terms.
Lemma 9.8. Consider ABC. Take a point D on the line BC.
AD ⊥ BC just when |AB|2 − |AC|2 = |DB|2 − |DC|2
Proof.
=⇒ Assume AD ⊥ BC. We prove the relation above. Indeed, D̂ = 90◦ . By Pythagoras’ theorem in ADB and ADC we have:
|AD|2 + |DB|2 = |AB|2 ,
|AD|2 + |DC|2 = |AC|2 .
After subtracting the two equations term by term and canceling out |AD|:
|DB|2 − |DC|2 = |AB|2 − |AC|2 .
⇐= Assume |DB|2 − |DC|2 = |AB|2 − |AC|2 . We’ll prove AD ⊥ BC. Indeed, assume
D′ is the point on BC such that AD′ ⊥ BC. Then by part I,
|D′ B|2 − |D′ C|2 = |AB|2 − |AC|2 , while
|DB|2 − |DC|2 = |AB|2 − |AC|2
22
by assumption. Hence |D′ B|2 − |D′ C|2 = |DB|2 − |DC|2 , or equivalently,
(9.1)
(|D′ B| − |D′ C|)(|D′ B| + |D′ C|) = (|DB| − |DC|)(|DB| + |DC|).
But |D′ B| + |D′ C| = |DB| + |DC| = |BC| if both D and D′ are inside the
segment BC, (which happens when ABC is acute-angled), or |D′ B| − |D′ C| =
|DB| − |DC| = ±|BC| if D and D′ are outside the segment BC, both on the
same side of the vertices B and C (which is the case when B̂ > 90◦ or Ĉ > 90◦ ).
After cancelling out the equal factors in equation (9.1), we have both
|D′ B| + |D′ C| = |DB| + |DC| and
|D′ B| − |D′ C| = |DB| − |DC|.
which added/subtracted yield |D′ B| = |DB| and |D′ C| = |DC|, hence D = D′ .
We note that it would be impossible for one of D, D′ to be inside the segment
BC and the other outside. Indeed, if for example D′ is outside and D inside, then
|D′ B| − |D′ C| = |DB| + |DC| = |BC|. After canceling these factors in equation
(9.1), we’d get |D′ B| + |D′ C| = |DB| − |DC| which is impossible as the positions
of D′ and D imply |D′ B| + |D′ C| ≥ |D′ B| > |BC| > |DB| > |DB| − |DC|.
In the right-angled triangle ADB with D̂ = 90◦ we define
cos B :=
|BD|
|AD|
and sin B :=
.
|AB|
|AB|
Note: cos B and sin B thus defined seems to depend on the choice of the triangle ABD.
We will prove later that they depend in fact only on the measure of the angle B̂.
Corollary 9.9 (The cosine formula). Consider ABC with the side lengths denoted by
|AB| = c, |AC| = b and |BC| = a. Let D be the point on BC such that AD ⊥ BC. Then
cos B =
a2 + c2 − b2
2ac
Proof. Consider the case when ABC is acute angled. From Lemma 9.8 we have |DB|2 −
|DC|2 = |AB|2 − |AC|2 = c2 − b2 . But |DC| = |BC| − |DB| = a − |DB|. Substituting
this in the equation above we get: |DB|2 − (a − |DB|)2 = c2 − b2 .. Solving for |DB| we
2
2
2
get |DB| = a +c2a−b .
Corollary 9.10 (Heron’s formula for the area of a triangle). Consider ABC with the side
lengths denoted by |AB| = c, |AC| = b and |BC| = a. Then
Area ABC =
p
p(p − a)(p − b)(p − c)
where p = 21 (a + b + c) is the semiperimeter of ABC.
23
Proof. Continuing with the calculations from the previous proof, we apply Pythagoras’
2
2
2
theorem in ADB with |AB| = c and |DB| = a +b2a−c to get
|AD| =
=
=
p
|AB|2 − |DB|2 ,
r
4a2 c2 − (a2 + c2 − b2 )2
,
4a2
r
c2 −
(a2 + c2 − b2 )2
,
4a2
p
(2ac − a2 − c2 + b2 )(2ac + a2 + c2 − b2 )
,
2a
p
(b2 − (a − c)2 )((a + c)2 − b2 )
=
,
2a
p
(b + a − c)(b − a + c)(a + c − b)(a + c + b)
,
=
2a
p
4 p(p − a)(p − b)(p − c)
=
2a
=
As |AD| is the height in ABC with basis |BC| = a, we get the formula for the area as
above.
Theorem 9.11 (Altitude as a geometric locus). Let ABC be a triangle and H a point
in the plane. Then H is on the altitude from A to BC just when AH ⊥ BC which occurs
just when |AB|2 − |AC|2 = |HB|2 − |HC|2 .
Proof. Let D be the intersection of the lines AH and BC.
=⇒ Applying Lemma 9.8 to AD ⊥ BC and then HD ⊥ BC we get
|AB|2 − |AC|2 = |DB|2 − |DC|2 = |HB|2 − |HC|2
⇐= Let D, D′ be points on BC such that AD ⊥ BC and then HD′ ⊥ BC. Thus by
Lemma 9.8,
|AB|2 − |AC|2 = |DB|2 − |DC|2 and |HB|2 − |HC|2 = |D′ B|2 − |D′ C|2 .
We assumed |AB|2 − |AC|2 = |HB|2 − |HC|2 , and so |DB|2 − |DC|2 = |D′ B|2 −
|D′ C|2 which as before implies D = D′ . Thus the lines AD ⊥ BC and HD′ ⊥
BC have a common point D = D′ . But only one perpendicular to BC can be
constructed from the point D and hence A, D and H must be collinear, AH ⊥ BC.
An alternate proof for the Orthocenter Theorem: Note that two altitudes BE and
CF must intersect at a point. Indeed, assume BE k CF . Then BE ⊥ AC would imply
CF ⊥ AC whereas from definition CF ⊥ AB. But then the triangle formed by the lines
AB, AC and CF would have two right angles, which is impossible.
We denote by H the intersection of the two altitudes BE and CF . We will prove
that H is also on a point on the altitude from A. Indeed, we can apply the Altitude as
geometric locus Theorem:
BH ⊥ AC just when |BA|2 − |BC|2 = |HA|2 − |HC|2
CH ⊥ AB just when |CB|2 − |CA|2 = |HB|2 − |HA|2 .
Subtract: |BA|2 − |CA|2 = |HB|2 − |HC|2 ,
which implies AH ⊥ BC.
24
9.5. Median. The median from the vertex A of a triangle ABC is the line joining A with
the midpoint of BC. The midpoint C ′ of the segment AB is joined to the midpoint B ′ of
AC by the midline B ′ C ′ of the triangle ABC.
Proposition 9.12 (Midlines). Let ABC be a triangle and consider the midpoint M of
the segment AB and the midpoint N of AC.
a) M N kBC and |M N | = |BC|/2.
b) Let G be the point of intersection of BN and CM . Then |BG| = 2|GN | = 23 |BN |
and |CG| = 2|GM | = 23 |CM |.
Proof. a): Extend M N by M P of equal length. By the Theorem on parallelograms (D),
AM CP is a parallelogram. This implies that CP kAB and |CP | = |AM | = |M B|. Thus
by the Theorem on parallelograms (B), BM P C is a parallelogram too, which implies
M P kBC and 2|M N | = |M P | = |BC|. This proves a).
Ab
Mb
b
N
b
P
G
b
R
b
b
S
Bb
b
C
b) Draw the midpoints R and S of the segments BG and CG, respectively. Then N S is
a midline in CGA and so by a),
N SkAG and |AG| = 2|N S|.
Similarly, M R is a midline in BGA and so by a),
M RkAG and |AG| = 2|M R|.
From the previous two observations, N SkM R and |N S| = |M R|. By the Theorem on
parallelograms (B), M N SR is a parallelogram, and so the diagonals intersect each other
at midpoints. Thus |M G| = |GS| = |SC| = |CG|/2 (because S was constructed as the
midpoint of the segment CG), and similarly |N G| = |GR| = |RG| = |BG|/2.
Corollary 9.13. Take a triangle ABC and split each side in half, drawing all 3 midlines.
Then these midlines split the triangle into 4 congruent triangles.
25
b
b
b
b
b
b
Proof. Each side of the triangle bound by the midlines is parallel to and half the size of
the obvious side of ABC, so all four small triangles are congruent by SSS.
Lemma 9.14. If ABC and A′ B ′ C ′ have the same angles at B and B ′ , and at C and C ′ ,
and if |B ′ C ′ | = 2|BC|, then every side of A′ B ′ C ′ is double the length of the corresponding
side of ABC.
Proof. Chop up A′ B ′ C ′ by midlines as above, to find 4 triangles, all congruent to ABC
by ASA.
Theorem 9.15 (The centroid of a triangle). All of the medians of a triangle intersect at
a point G, the centroid of the triangle.
Proof. Let Q be the midpoint of BC. With the notations from the Proposition on Midlines,
AQ, BN and CM are the median in the triangle ABC, and G is the point of intersection
of BN and CM . Assume that BN and AQ intersect at another point G′ . We apply the
Proposition on Midlines, b) in two cases:
• to G as the point of intersection of BN and CM so |BG| = 23 |BN |.
• to G′ as the point of intersection of BN and AQ so |BG′ | = 23 |BN |.
From here it follows that G = G′ , since |BG| = |BG′ | and G, G′ are both interior
points of BN . Thus AQ, BN and CM all intersect at G.
Theorem 9.16 (Median as a geometric locus). A point G in the interior of a triangle
ABC is on one of its medians AM just when Area ABG = Area ACG.
Proof.
=⇒ Note that Area ABM = Area ACM as the triangles have equal bases and the
same height. Similarly, Area GBM = Area GCM . Subtracting the two equations
above yields Area ABG = Area ACG.
⇐= As the two triangles have a common side AG, it follows that their heights |BE|
and |CF | are equal.
26
b
A
b
G
90◦b E
B
b
b
M
b
b
90
C
F
◦
Then BEM ≡ CF M by AAS, as they have:
• |BE| = |CF |
\ = CF
\
• BEM
M = 90◦
\
\
• BM
E = CM
F as opposite angles.
Hence, |BM | = |CM | and so AM is median.
An alternate proof for the theorem of the centroid:
Proof. Two medians BN and CP of ABC will always intersect at a point G, as they are
both in the interior of ABC. We will use the characterization of medians as geometric
locus to prove that G is also a point on the median AM . Indeed, a point G of BN is a
median just when Area BAG = Area BCG and a point G of CP is a median just when
Area BCG = Area CAG. Hence Area BAG = Area CAG, and so G is also a point on the
median AM .
Theorem 9.17 (Euler’s line). The orthocentre H, circumcentre O, and centroid G of any
triangle ABC are collinear and satisfy |HG| = 2|HO|.
27
A
b
b
b
b
H
b
B
b
G
O
b
b
M
b
b
C
A′
Proof. Let M be the midpoint of the segment BC and let AA′ be the diameter of the
circumcircle of ABC amd similarly construct a diameter BB ′ . Then AB ′ A′ B is bisected by
its diagonals, so is a rectangle. Hence A′ B ⊥ AB. Because CH is an altitude, CH ⊥ AB
so A′ B k CH. By the same reasoning, replacing B with C and vice versa, A′ C ⊥ AC
and BH ⊥ AC, so A′ C k BH. Put these two last sentences together: BHCA′ is a
parallelogram. Hence M is the midpoint of segment HA′ . The point G is on AM , one
of the medians of AHA′ , so a candidate for the centroid of AHA′ , and cuts that median
AM in ratio 2 : 1. Hence G is the centroid of AHA′ . As such, G is on the median HO
and |HG| = 2|HO| by the Theorem of the Centroid.
10. Circles
We will denote by Cr O a circle of center O and radius r.
10.1. Secants and tangents.
Lemma 10.1. Let Cr O be a circle of centre O and radius r and let A, B be two points
on the circle. Then the perpendicular from O to AB intersects the segment AB at its
midpoint P , and
1
|OP |2 = r2 − |AB|2 .
4
Proof. Let P denote the point of intersection of AB with the perpendicular from O to
AB. Then by applying Pythagora’s theorem in the right angled triangles OP A and OP B
we get
|AP |2 = |OA|2 − |OP |2 = |OB|2 − |OP |2 = |BP |2 .
As |AP | =
lemma.
1
|AB|,
2
the equation above can be rearranged like in the conclusion of the
A secant is a line which intersects the circle at two points. A tangent is a line which
intersects the circle at exactly one point. We will think of the tangent as the limit of a
sequence of secants passing through a fixed point M , and moving gradually further away
from the center O of the circle.
28
Let M be a point outside a line d, and let P be a point on d such that M P ⊥ d. We
say that P is the projection of M on the line d. If M is on d, we then define the projection
of M on d to be M .
Lemma 10.2. Given any point M and line AB, there is a unique projection of M on
AB.
Proof. Suppose that there are two projections of a point M on a line d: P and P ′ . The
triangle P M P ′ has two right angles, a contraction. So the projection is unique if it exists.
Construct a perpendicular bisector to AB; call it ℓ. Find the parallel to ℓ through M ;
call it ℓ′ . Let Q be the point at which ℓ is perpendicular to AB. If ℓ′ is parallel to AB,
then ℓ′ has two parallels through Q, perpendicular to one another, so not equal to one
another, a contradiction. Hence ℓ′ meets AB, at some point P , a projection.
Theorem 10.3. Let M be a point not situated on a circle C of center O, and let M P be
tangent to C , where P is a point on the circle. Then OP ⊥ M P .
Proof. Suppose that OP is not perpendicular to M P . Project O perpendicularly to M P ,
so at some point Q 6= P . By Pythagorean theorem, |OQ| < |OP |, so Q lies
p inside C . Pick
the point R on P Q, in the direction from P to Q, of distance |RQ| = |OP |2 − |OQ|2 .
By Pythagorean theorem, |OR| = |OP |, so R lies on C .
Conversely, we have:
Theorem 10.4. Let M be a point not situated on a circle C with center O, and let P is
a point on the circle. Assume OP ⊥ M P . Then M P is tangent to the circle C .
Proof. We want to show that M P is tangent to the circle C , i.e., by definition, that
P is the unique point of intersection of M P with C . Uniqueness is most often proven
by contradiction, so we will assume that there exists another point of intersection P ′ .
Then since P and P ′ are both on the circle, the triangle OP P ′ is isosceles and as such,
Pˆ′ = P̂ = 90◦ (since OP ⊥ M P ). But the angles in OP P ′ can’t sum up to more than
180◦ – thus our assumption of the existence of P ′ must have been wrong.
Lemma 10.5. Let M be a point not situated on a circle C , and let M P and M P ′ be
tangents to C , where P and P ′ are points on the circle. Then |M P | = |M P ′ |.
Proof. This follows by applying the Pythagorean theorem in triangles OM P and OM P ′ .
Theorem 10.6. Let C1 = Cr1 O1 and C2 = Cr2 O2 be two circles intersecting at two points
A and B. Then O1 O2 ⊥ AB and O1 O2 passes through the midpoint of the segment AB.
\
\
Proof. AO1 O2 ≡ BO1 O2 (case SSS) just when AO
2 O1 = BO
2 O1 just when O1 O2 is angle
bisector in the isosceles triangle O2 AB with |O2 A| = |O2 B| just when O1 O2 ⊥ AB and
O1 O2 passes through the midpoint of the segment AB (as O1 O2 must also be perpendicular
bisector in triangle O2 AB).
Two circles are tangent to each other if they intersect at only one point. Just like with
a circle and a line, we consider tangency of two circles as the limit position of a sequence
of secant circles. Hence the following theorem.
Theorem 10.7. Let Cr1 O1 and Cr2 O2 be two circles tangent to each other at the point
P . Then O1 , O2 and P are collinear.
Proof. Reflect across the line O1 O2 , preserving both circles, and hence fixing the point P .
But the fixed points of the reflection form the line O1 O2 .
29
>
\
10.2. Arcs and angles. In a circle Cr O, the arc AB is the angle AOB.
Lemma 10.8 (Angle on a circle). Take a circle with centre O and let A, B, C be three
>
points on that circle. Let BC denote the arc which does not contain A. Then
>
\
\ = BOC = BC .
BAC
2
2
Proof. Let D be the point on the circle such that A, O and D are collinear. Then
\ = 2OAB,
\ as exterior angle of the isosceles triangle OAB, and
• BOD
\ = 2OAC,
\ as exterior angle of the isosceles triangle OAC.
• COD
\ we add the equations above term by term, and
If O is in the interior of the angle BAC,
\
O is outside of the angle BAC, we subtract the equations above term by term. In both
\ = 2BAC.
\
cases, we obtain BOD
Corollary 10.9 (Internal and external angles). Let C be a circle of center O. Let A, B,
C, D be points on C and P the intersection point of AB and CD, which we suppose lies
in the interior of C .
C
b
B
b
P
b
Q
b
b
D
b
A
Let Q be the intersection point of the lines AD and BC. Then
>
1 > >
\ = 1 (>
\
AC − BD).
AP
C = (AC + BD) and AQC
2
2
\
Proof. AP
C is exterior angle for P BC so
\
\ + DCB,
\
AP
C = ABC
\
\
=P
BC + P
CB,
1 > >
= (AC + BD)
2
\ is exterior angle for QBA so AQB
\ = ABC
\ − BAD
\ =
(angles on the circle). ABC
>
1 >
(AC − BD).
2
Lemma 10.10. Let A, B, C be three points on a circle. Also, let BD be tangent to the
circle, with D on the same side of AB as C. Then
\ = DBC.
\
BAC
The proof is left as an exercise.
A quadrilateral ABCD is cyclic if all its vertices are on a circle.
Lemma 10.11 (Isosceles trapezoid in a circle). Let ABCD be a cyclic quadrilateral. The
following are equivalent:
a) |AD| = |BC|
> >
b) AD = BC
c) AB k CD
30
\ = BOC
\ just
Proof. a) |AD| = |BC| just when OAD ≡ OBC (case SSS) just when AOD
>
>
> >
\ = AD = BC = BDC
\ (angles on the circle) just when
when b) AD = BC just when ABD
2
2
c) AB k CD.
Lemma 10.12 (Rectangle in a circle). Let ABCD be a quadrilateral whose vertices are
on a circle Cr O. The following are equivalent:
a) ABCD is a rectangle.
b) AC and BD are diameters of the circle (i.e., A, O, D are collinear, and B, O, C are
collinear).
Proof. Exercise.
Theorem 10.13. Let ABCD be a quadrilateral. The following are equivalent:
a) The quadrilateral ABCD is cyclic.
\ = ACD.
\ (The angle formed by a diagonal with a side is equal with that
b) ABD
formed by the other diagonal with the opposite side).
\ + ADC
\ = 180◦ (The sum of two opposite angles is 180◦ ).
c) ABC
b
B
b
C
Ab
b
D
Proof.
a) =⇒ b) and c) On the circle C containing the vertices A, B, C, D, we have
>
\ = ACD
\ = AD ,
ABD
2
> >
◦
\ + ADC
\ = ADC + ABC = 360 = 180◦ .
ABC
2
2
>
Here ADC denotes the arc bounded by A and C and containing the point D,
>
while ABC denotes the arc bounded by A and C and containing the point B.
b) =⇒ a) Let C be the circle containing the points A, B, C. This is the circle whose centre
is the circumcenter O of the triangle ABC (the intersection of the perpendicular
bisectors), and whose radius is |OA|. We would like to prove that D is also a
point on the circle C . Proof by contradiction: assume D is not on the circle C .
Let C intersect the line CD at the point E, the line BD at the point F .
31
b
b
B
Ab
A
b
b
C
b
C
D
b
b
B
b
b
b
F
F
E
b
E
D
We have
>
>
\ = AF while ACD
\ = AE ,
ABD
2
2
\ = ACD,
\ and so by the
as angles on the circle. By assumption, we know ABD
> >
equations above, AF = AE.
>
However, if D is outside the circle C , then F is inside the arc AE and so
> >
AF < AE. Contradiction.
>
However, if D is inside the circle C , then E is inside the arc AF and so
> >
AE < AF . Contradiction.
c) =⇒ a) Similar with the previous part. Let C be the circle containing the points A, B,
C. We would like to prove that D is also a point on the circle C . Proof by
contradiction: assuming D is not on the circle C , let C intersect the line CD at
the point E, the line AD at the point L. We have
>
\ = ALC ,
ABC
2
as angle on the circle, and
> >
\ = ABC ± EF L ,
ADC
2
as angle which is either internal, or external to the circle. By assumption, we
\ + ADC
\ = 180◦ , and so by the equations above,
know ABC
> > >
ALC + ABC ± EF L = 360◦ .
However,
> >
ALC + ABC = 360◦
>
as they span the entire circle, so EF L = 0, meaning that L = E. But this would
mean that the lines CD and AD intersect the circle at the same point E = L. As
the intersection of AD and CD is D, we must have D = E = L.
Example 10.14. Let H be the orthocentre of ABC, and let D′ denote the symmetric of
H through BC. Then D′ is a point on the circumcentre of ABC.
32
b
A
b
b
E
b
b
b
B
b
O
H
D
b
C
D′
′C =
\
Proof. BC is the perpendicular bisector of HD′ =⇒ CDH ≡ CDD′ . Then AD
◦
′
\
\
\
DHC = 90 − HCD = ABC so the quadrilateral ABD C is cyclic.
(We used HD ⊥ BC and CH ⊥ AB. )
11. Similar triangles
′
′
′
Triangles ABC and A B C are similar if their respective angles are equal: Â = Â′ ,
B̂ = B̂ ′ , Ĉ = Ĉ ′ . we write
ABC ∼ A′ B ′ C ′
Theorem 11.1. If ABC ∼ A′ B ′ C ′ then
|AC|
|BC|
|AB|
= ′ ′ =
.
|A′ B ′ |
|A C |
|B ′ C ′ |
As an application, sin α and cos α as defined in trigonometry are independent of the
choice of the triangle in which they are computed.
The most common situations when two similar triangles arise are the following:
Theorem 11.2 (Parallel lines III). Given two triangles ABC and AB ′ C ′ sharing a vertex
A, if B ′ lies on AB and C ′ lies on AC, then BC k B ′ C ′ just when ABC ∼ AB ′ C ′ .
b
B′
B
b
b
A
b
C′
b
C
Theorem 11.3 (Anti-parallel lines). If B, C, C ′ , B ′ are on the same circle and BC ′ intersects CB ′ at A then ABC ∼ AB ′ C ′ ; BC and B ′ C ′ are anti-parallel.
33
b
b
B′
b
A
b
B
b
b
C
A
B
Bb
b
b
b
B′
b
C′
b
b
C
b
C′
b
C′
In particular, we can solve the following question: Given a circle Cr O and a point P ,
how can we describe how far the point is from the circle?
The power of a point P with respect to a circle is the product of the lengths of the
segments made by the point P on any chord passing through P .
Theorem 11.4. The power of a point P with respect to a circle Cr O does not depend on
the chord on which it is calculated:
|P A| · |P B| = |P A′ | · |P B ′ | = ±(|P O|2 − r2 ),
+ if P is outside the circle and − if P is inside the circle.
b
b
B
Ab ′
b
A
B′
A
b
B
Ab
P
b
O
b
P
b
b
A′
b
O
b
B′
B′
Proof. Due to the equal angles on the circle, P AA′ ∼ P B ′ B so that
|P A|
|P A′ |
=
|P B ′ |
|P B|
and thus |P A||P B| = |P A′ ||P B ′ | = ±(|OP | − r)(|OP | + r)..
C
34
12. The nine-point circle
E′
A
b
b
B′
C′
b
b
M′
E
b
b
N
P
b
b
O
b
G
b
O
′
b
F′
b
F
b
H
b
N′
b
b
P′
D
b
b
b
b
B
M
b
D
b
′
A′
C
35
Can you guess the significance of each point? If you were to connect all labeled points,
could you find
• at least 19 segments whose midpoints are labeled in?
• at least 9 perpendicular bisectors?
• at least 15 parallelograms which are not rectangles?
• at least 9 isosceles trapezoids?
• at least 6 diameters and 9 rectangles?
• at least 9 pairs of similar triangles sharing H as common vertex?
• at least 12 pairs of similar triangles sharing A as common vertex?
• at least 8 triangles having G as centroid?
13. Menelaus and Ceva theorems
Theorem 13.1 (Menelaus of Alexandria c. 70 – 140 CE) ). Let C̄, B̄ be points inside the
segments AB, and CA, and let Ā be a point on the line BC, outside of the segment BC.
Ā, B̄, C̄ are collinear if and only if
|AC̄| |B Ā| |C B̄|
=1
|C̄B| |ĀC| |B̄A|
A
b
C̄
b
B̄
b
B
b
b
Ā
b
C
Proof. If the points are collinear, then to prove relation: Let CD k AB.
b
C̄
A
b
b
B̄
b
B
b
b
D
b
C
Ā
|C̄B|
Ā|
B̄|
= |CD|
= |CD|
.
Then multiply |B
with |C
|ĀC|
|B̄A|
|AC̄|
If we know the relation, assume the points are not collinear, let C ′ be the intersection
of lines ĀB̄ and AB. It has to lie inside the segment AB just like C̄. Then by the previous
argument,
|AA′ | |B Ā| |C B̄|
=1
|A′ B| |ĀC| |B̄A|
while by assumption
|AC̄| |B Ā| |C B̄|
=1
|C̄B| |ĀC| |B̄A|
hence
′
|AC ′ |
|C ′ B|
C = C̄.
=
|AC̄|
|C̄B|
so
|AC ′ |
|AB|
=
|AC ′ |
|AC ′ |+|C ′ B|
=
|AC̄|
|AC̄|+|C̄B|
=
|AC ′ |
|AB|
so |AC̄| = |AC ′ | thus
36
The following theorem is known as Ceva’s theorem (after Giovanni Ceva, December 7,
1647 – June 15, 1734), but was actually proven by the Spanish Arab mathematician and
aristocrat Yusuf al-Mu’taman ibn Hud, who was king of Zaragossa from 1081 to 1085. His
palace is depicted on the frontispiece of our lecture notes.
Theorem 13.2 (“Ceva’s” theorem). Let C̄, Ā, B̄ be points inside the segments AB, BC
and CA. The lines AĀ, B B̄ and C C̄ are concurrent if and only if
|AC̄| |B Ā| |C B̄|
= 1.
|C̄B| |ĀC| |B̄A|
Proof.
a) We assume the lines AĀ, B B̄ and C C̄ are concurrent. We apply Menelaus’ Theorem
twice: once for triangle AB Ā crossed by line C C̄ and then for triangle AC Ā crossed
by line B B̄. Multiplying the two ensuing relations yields the formula above.
b) We assume the formula above. We let A′ be the intersection point of lines B B̄ and
C C̄. Let A′′ denote the intersection of line AA′ with BC. From Part I, we get:
|AC̄| |BA′′ | |C B̄|
= 1.
|C̄B| |A′′ C| |B̄A|
From assumption, we have
|AC̄| |B Ā| |C B̄|
= 1.
|C̄B| |ĀC| |B̄A|
Together, these yield
|BA′′ |
|B Ā|
.
=
|A′′ C|
|ĀC|
Since both Ā and A′′ lie in the same segment BC, with the same ratios of distances,
Ā = A′′ .
References
[1] Michèle Audin, Geometry, Springer-Verlag, Heidelberg, 2003.
[2] H.S.M. Coxeter, Introduction to Geometry, 2ed., Wiley, 1969.
[3] Euclid, The Elements, on-line editions:
http://farside.ph.utexas.edu/euclid/Elements.pdf
http://www.gutenberg.org/files/21076/21076-pdf.pdf
[4] Geometry Package GeoGebra, http://www.geogebra.org
[5] Robin Hartshorne, Geometry: Euclid and Beyond, Springer-Verlag, Heidelberg, 2000.
[6] Silvio Levy, preface to Flavours of Geometry, MSRI, Berkeley, 1997.
Index
acute angle, 4
acute-angled triangle, 5
al-Mu’taman ibn Hud, Yusuf, 36
altitude, 20
theorem, 23
angle, 4
acute, 4
alternate, 8
corresponding, 8
interior consecutive, 8
obtuse, 4
angle bisector, 17
theorem, 17
angle-side-angle, 11
anti-parallel, 32
arc, 29
area, 15
ASA, 11
axiom, 3
hypotenuse, 5
hypothesis, 3
in between, 6
incentre, 18
theorem, 18
incircle, 19
interior, 6
interior angle, 5
isosceles, 5
isosceles triangle
theorem, 11
leg, 4
lemma, 3
line
parallel, 5
locus, 17
mathematical theory, 3
measure, 6
median, 24
theorem, 25
Menelaus theorem, 35
midline, 24
bisector
angle, 17
centre, 5
centroid, 25
theorem, 25
Ceva theorem, 36
circle, 5
nine-point, 34
circumcentre, 19
circumcircle, 19
circumference, 5
cirumcentre
theorem, 19
collinear, 4
complement, 5
conclusion, 3
concurrent, 4
concyclic, 5
congruent, 6, 10
corollary, 3
cosine, 22
cyclic, 29
nine-point circle, 34
obtuse angle, 4
obtuse-angled triangle, 5
orthocenter, 20
orthocentre
theorem, 20
parallel, 7
parallel line, 5
parallelogram, 13
pentagon, 5
perpendicular, 4
perpendicular bisector, 19
theorem, 19
polygon, 5
regular, 10
power, 33
projection, 28
proof, 3
proposition, 3
Pythagorean theorem, 16
definition, 3
diameter, 5
distance
from point to line, 17
quadrilateral, 5
equilateral, 5
Euler, 26
Euler’s line, 26
exterior angle, 5
radius, 5
rays, 4
rectangle, 14, 35
theorem, 14
rhombus, 15
rhombus theorem, 15
right angle, 4
geometric locus, 17
half-line, 4
hexagon, 5
37
38
right-angled triangle, 5
rigid motion, 5
SAS, 11
scalene, 5
secant, 27
segment, 4
side, 5
side-angle-side, 11
side-side-side, 12
similar triangles, 32
sine, 22
smaller, 6
SSS, 12
sum, 4
sum of angles in a triangle, 9
supplement, 4
tangent, 19, 27
circles, 28
tessellation, 10
theorem, 3
altitude, 23
angle bisector, 17
angle-side-angle, 11
centroid, 25
Ceva, 36
circumcentre, 19
incentre, 18
isosceles triangle, 11
median, 25
Menelaus, 35
orthocentre, 20
parallel lines I, 7
parallel lines II, 8
parallel lines III, 32
parallelogram, 13
perpendicular bisector, 19
Pythagorean, 16
rectangle, 14
rhombus, 15
side-angle-side, 11
side-side-side, 12
sum of angles in a triangle, 9
tiling, 10
trapezoid, 29, 35
triangle, 5
vertex, 4
Zaragossa, 36