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RAMON GONZALEZ CALVET
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Proof ⇐ Being F a point on the line AB, D a point on the line BC and E a point on the line CA
we have:

F = a A + (1 ’ a) B D = b B + (1 ’ b) C E = c C + (1 ’ c) A

with a, b and c real. Then:

AF = (1 ’ a) AB BD = (1 ’ b) BC
FB = a AB

CE = (1 ’ c) CA
DC = b BC EA = c CA

That the product of these segments is equal to ’1 implies that:

(1 ’ a )(1 ’ b )(1 ’ c ) (1 ’ a )(1 ’ b )
= ’1 ’ c=
1’ a ’ b
abc

The substitution of c in the expression for E gives:

(1 ’ a )(1 ’ b ) ab
E= C’ A
1’ a ’ b 1’ a ’ b

1’ a
[(1 ’ b) C + b B] ’ b [a A + (1 ’ a ) B]
=
1’ a ’ b 1’ a ’ b

1’ a b
= D’ F
1’ a ’ b 1’ a ’ b


That is, the point E belongs to the line FD, in other words, the point D, E and F are
aligned, which is the prove:

1’ a
with d =
E=dD+(1’d)F
1’ a ’ b


The theorem of Ceva
Figure 7 5
Given a generic triangle ABC, we draw
three segments AD, BE and CF from each vertex
to a point of the opposite side (figure 7.5). The
three segments meet in a unique point if and
only if:

BD DC ’1 CE EA ’1 AF FB ’1 = 1

In order to prove the theorem, we
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 65


enlarge the segments BE and CF to touch the line which is parallel to the side BC and passes
through A. Let us denote the intersection points by M and N respectively. Then the triangle BDO
is directly similar to the triangle MAO, and the triangle CDO is also directly similar to the
triangle NAO. Therefore:

BD DO ’1 = MA AO ’1

DO DC ’1 = AO AN ’1

The multiplication of both equalities gives:

BD DC ’1 = MA AN ’1

Analogously, the triangles MEA and CEB are similar as soon as NFA and BFC. Hence:

CE EA ’1 = BC AM ’1 AF FB ’1 = AN BC ’1

The product of the three equalities yields:

BD DC ’1 CE EA ’1 AF FB ’1 = MA AN ’1 BC AM ’1 AN BC ’1 = 1

The sufficiency of this condition is proved in the following way: let O be the point of
intersection of BE and CF, and D' the point of intersection of the line AO with the side BC. Let
us suppose that the former equality is fulfilled. Then:

BD' D'C ’1 = BD DC ’1

Since both D and D' lie on the line BC, it follows that D = D'.


Homothety and single ratio

A homothety with centre O and ratio k is the geometric transformation which dilates the
distance from O to any point A in a factor k:
Figure 7.6
OA' = OA k

O is the unique invariant point of the
homothety. If k is real number, the homothety is
said to be simple (figure 7.6), otherwise is called
composite. If k is a complex number, it may always
be factorised in a product of the modulus and a
unitary complex number:

k = ¦k¦ z with ¦z¦ = 1

The modulus of k is the ratio of a simple homothety with centre O, and z indicates an
RAMON GONZALEZ CALVET
66


additional rotation with the same centre. Then, a composite homothety is equivalent to a simple
homothety followed by a rotation. Direct similarities and homotheties are different names for
the same transformations (exercise 7.5). Also homotheties with ¦k¦=1 and displacements are
equivalent.
The single ratio of three points A, B, and C is defined as:

( A B C ) = AB AC ’1

The single ratio is a real number when the three points are aligned, and a complex
number in the other case. Under a homothety, the single ratio of any three points remains
invariant. Let us prove this. Because any vector is dilated and rotated by a factor k, we have:

A'B' = OB' ’ OA' = OB k ’ OA k = AB k

A'C' ’1 = k ’1 AC ’1

A'C' = AC k

( A' B' C' ) = AB k k ’1 AC ’1 = AB AC ’1 = ( A B C )

If the single ratio is invariant for a geometric transformation, then it transforms triangles
into directly similar triangles and hence also polygons into similar polygons:
’1 ’1 ’1
A'C' = AB ’1 A'B'
’ AB AC ’ AC
( A B C ) = ( A' B' C' ) = A'B' A'C'

Therefore, the homothety always transforms triangles into directly similar triangles as
shown in the figure 7.6. It follows immediately that the homothety preserves the angles between
lines. It is the simplest case of conformal transformations, the geometric transformations which
preserve the angles between lines. Since a line
may also be transformed into a curve, and for
the general case a curve into a curve, the
conformal transformations are those which
preserve the angle between any pair of curves,
that is, the angle between the tangent lines to
both curves at the intersection point. A
transformation is directly or oppositely Figure 7.7
conformal whether it preserves or changes the
sense of the angle between two curves (figure
7.7). This condition is equivalent to the
conservation of the single ratio of any three
points at the limit of accumulation:

(A B C ) = ( A' B' C' )
lim lim directly conformal
B, C ’ A B' , C' ’ A'

(A B C ) = ( A' B' C' ) *
lim lim oppositely conformal
B, C ’ A B' , C' ’ A'
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 67




Exercises

7.1 Let T be the triangle with vertices (0, 0), (2, 0), (0, 1) and T' that with vertices (2,0), (5, 1)
and (4, 2). Find which vertices are homologous and calculate the similarity ratio. Which is
the size ratio? Which is the angle of rotation of the triangle T' with respect to T?

7.2 The altitude perpendicular to the hypotenuse divides a right triangle in two smaller right
triangles. Show that they are similar and deduce the Pythagorean theorem.

7.3 Every triangle ABC with not vanishing area has a circumscribed circle. The line tangent to
this circle at the point B cuts the line AC at the point M. Prove that:

MA MC ’1 = AB2 BC ’2

7.4 Let ABC be an equilateral triangle inscribed inside a circle. If P is any point on the arc BC,
show that ¦PA¦ = ¦PB¦ + ¦PC¦.

7.5 Given two directly similar triangles ABC and A'B'C', show that the centre O of the
homothety that transform one triangle into another is equal to:

O = A ’ AA' ( 1 ’ AB ’1 A'B' ) ’1

7.6 Draw any line passing through a fixed point P which cuts a given circle. Let Q and Q' be the
intersection points of the line and the circle. Show that the product PQ PQ' is constant for
any line belonging to the pencil of lines of P.

7.7 Let a triangle have sides a, b and c. The bisector of the angle formed by the sides a and b
divides the side c in two parts m and n. If m is adjacent to a and n to b respectively, prove
that the following proportion is fulfilled:

m n
=
a b
RAMON GONZALEZ CALVET
68

8. PROPERTIES OF TRIANGLES

Area of a triangle

Since the area of a parallelogram is obtained as the outer product of two
consecutive sides (taken as vectors, of course), the area of a triangle is the half of the outer
product of any two of its three sides:

1 1 1
aPQR = PQ § PR = QR § QP = RP § RQ
2 2 2

If the vertices P, Q and R are counterclockwise oriented, the area is a positive
imaginary number. Otherwise, the area is a
negative imaginary number. Note that the
fundamental concept in geometry is the
Figure 8.1
oriented area1. The modulus of the area
may be useful in the current life but is
insufficient for geometry. From now on I
shall only regard oriented areas.
Writing the segments of the former
equation as differences of points we arrive
to:

1
(P § Q + Q § R + R § P )
aPQR =
2

which is a symmetric expression under cyclic permutation of the vertices. The position
vector P goes from an arbitrary origin of coordinates to the point P. Then, P § Q is the
double of the area of the triangle OPQ. Analogously Q § R is the double of the area of the
triangle OQR and R § P is the double of the area of the triangle ORP. Therefore the
former expression is equal to (figure 8.1):

aPQR = aOPQ + aOQR + aORP

For the arrangement of points shown in the figure 8.1 the area of OPQ is positive,
and the areas of OQR and ORP are negative, that is, the areas of ORQ and OPR are
positive. Therefore, one would intuitively write, taking all the areas positive, that:

aPQR = aOPQ ’ aORQ ’ aOPR

When one considers oriented areas, the equalities are wholly general and
independent of the arrangement of the points2.



1
The integral of a function is also the oriented area of the region enclosed by the curve and the X-
axis.
2
About this topic, see A. M. Lopshitz, Cálculo de las áreas de figuras orientadas, Rubi±os-1860
(1994).
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 69

Medians and centroid

The medians are the segments going from each vertex to the midpoint of the
opposite side. Let us prove that the three medians meet in a unique point G called the
centroid (figure 8.2). Since G is a point on the median passing through P and (Q + R)/2:

Q+R
G = k P + (1 ’ k ) k real
2

G lies also on the median passing through Q and (P + R) / 2:

P+R
G = m Q + (1 ’ m ) m real
2

Equating both expressions we find:

Q+R P+R
k P + (1 ’ k ) = m Q + (1 ’ m )
2 2

« 1 m «1 k  « k m
P ¬k ’ + · + Q ¬ ’ ’ m· + R ¬’ + · = 0
2 2 2 2  2 2
 

A linear combination of independent points can vanish only if every coefficients are null,
a condition which leads to the following
system of equations:
Figure 8.2
± 1m
k ’ + =0
 22
1 k

 ’ ’m=0
2 2
 ’ k + m =0
22


The solution of this system of
equations is k =1/3 and m =1/3, indicating
that the intersection of both medians are
located at 1/3 distance from the midpoints. The substitution into the expression of G
gives:

P+Q + R
G=
3

This expression for the centroid is symmetric under permutation of the vertices. Therefore
the three medians meet at the same point G, the centroid3.

3
The medians are a special case of cevian lines (lines passing through a vertex and not
coinciding with the sides) and this statement is also proved by means of Ceva™s theorem.
RAMON GONZALEZ CALVET
70




Perpendicular bisectors and circumcentre

The three perpendicular bisectors of the sides of a triangle meet in a unique point
called the circumcentre, the centre of the circumscribed circle. Every point on the
perpendicular bisector of PQ is equidistant from P and Q. Analogously every point on the
perpendicular bisector of PR is equidistant from P and R. The intersection O of both
perpendicular bisectors is simultaneously equidistant from P, Q and R. Therefore O also
belongs to the perpendicular bisector of QR and the three bisectors meet at a unique point.
Since O is equally distant from the three vertices, it is the centre of the circumscribed
circle. Let us use this condition in order to
calculate the equation of the circumcentre:
Figure 8.3
OP2 = OQ2 = OR2 = d2

where d is the radius of the circumscribed
circle. Using the position vectors of each
point we have:

(P ’ O)2 = (Q ’ O)2 = (R ’ O)2

The first equality yields:

P2 ’ 2 P · O + O2 = Q2 ’ 2 Q · O + O2

By simplifying and arranging the terms containing O at the left hand side, we have:

2 ( Q ’ P ) · O = Q2 ’ P2

2 PQ · O = Q2 ’ P2

From the second equality we find an analogous result:

2 QR · O = R2 ’ Q2

Now we introduce the geometric product instead of the inner product in these equations:

PQ O + O PQ = Q2 ’ P2

QR O + O QR = R2 ’ Q2

By subtraction of the second equation multiplied on the right by PQ minus the first
equation multiplied on the left by QR, we obtain:

PQ QR O ’ O PQ QR = PQ R2 ’ PQ Q2 ’ Q2 QR + P2 QR

By using the permutative property on the left hand side and simplifying the right hand
side, we have:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 71


PQ QR O ’ QR PQ O = P2 QR + Q2 RP + R2 PQ

2 ( PQ § QR ) O = P2 QR + Q2 RP + R2 PQ

Finally, the multiplication by the inverse of the outer product on the left gives:

O = ( 2 PQ § QR ) ’1 ( P2 QR + Q2 RP + R2 PQ )

= ’ ( P2 QR + Q2 RP + R2 PQ ) ( 2 PQ § QR ) ’1

a formula able to calculate the coordinates of the circumcentre. For example, let us
calculate the centre of the circle passing through the points:

R = ( 4, ’2 )
P = ( 2, 2 ) Q = ( 3, 1 )

P2 = 8 Q2 = 10 R2 = 20

QR = R ’ Q = e1 ’ 3 e2 RP = P ’ R = ’ 2 e1 + 4 e2 PQ = Q ’ P = e1 ’ e2

2 PQ § QR = ’ 4 e12

e12
O = ’ ( 8 ( e1 ’ 3 e2 ) + 10 ( ’2 e1 + 4 e2 ) + 20 ( e1 ’ e2 ) )
4

= ’ e1 ’ 2 e2 = ( ’1, ’2 )

In order to deduce the radius of the circle, we take the vector OP:

OP = P ’ O = P + ( P2 QR + Q2 RP + R2 PQ ) ( 2 PQ § QR ) ’1

and extract the inverse of the area as a common factor:

OP = ( 2 P PQ § QR + P2 QR + Q2 RP + R2 PQ ) ( 2 PQ § QR ) ’1 =

= [ 2 P ( P § Q + Q § R + R § P ) + P2 QR + Q2 RP + R2 PQ ] ( 2 PQ § QR ) ’1 =

= [ P ( P Q ’ Q P + Q R ’ R Q + R P ’ P R ) + P2 ( R ’ Q ) + Q2 ( P ’ R ) +

+ R2 ( Q ’ P ) ] ( 2 PQ § QR ) ’1

The simplification gives:

OP = ( P Q R ’ P R Q + P R P ’ P Q P + Q2 P ’ Q2 R + R2 Q ’ R2 P ) ( 2 PQ § QR ) ’1


= ’ (Q ’ P) (R ’ Q) (P ’ R) ( 2 PQ § QR ) ’1 = ’ PQ QR RP ( 2 PQ § QR ) ’1
RAMON GONZALEZ CALVET
72

Analogously:

OQ = ’ QR RP PQ ( 2 PQ § QR ) ’1 OR = ’ RP PQ QR ( 2 PQ § QR ) ’1

The radius of the circumscribed circle is the length of any of these vectors:

PQ QR RP PQ QR RP
OP = = = =
2 PQ § QR 2 sin QRP 2 sin RPQ 2 sin PQR

where we find the law of sines.


Angle bisectors and incentre

The three bisector lines of the angles of a Figure 8.4
triangle meet in a unique point called the incentre.
Every point on the bisector of the angle with
vertex P is equidistant from the sides PQ and PR
(figure 8.4). Also every point on the angle
bisector of Q is equidistant from the sides QR and
QP. Hence its intersection I is simultaneously
equidistant from the three sides, that is, I is unique
and is the centre of the circle inscribed into the
triangle.
In order to calculate the equation of the
angle bisector passing through P, we take the sum
of the unitary vectors of both adjacent sides:

PQ PR QP QR
u= + v= +
PQ PR QP QR

The incentre I is the intersection of the angle bisector passing through P, whose
direction vector is u, and that passing through Q, with direction vector v:

I=P+ku=Q+mv k, m real

Arranging terms we find PQ as a linear combination of u and v:

k u ’ m v = Q ’ P = PQ

The coefficient k is:

PQ RP PQ § QR
PQ § v
k= =
u§v PQ § QR RP + QR § RP PQ + RP § PQ QR

Since all outer products are equal because they are the double of triangle area, this
expression is simplified:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 73

PQ RP
k=
RP + PQ + QR

Then, the centre of the circumscribed circle is:

« PQ PR 
PQ RP
¬ ·
I =P+ku=P+ +
PQ + QR + RP ¬ PQ PR ·
 

By taking common denominator and simplifying, we arrive at:

P QR + Q RP + R PQ
I=
QR + RP + PQ

For example, let us calculate the centre of the circle inscribed inside the triangle with
vertices:

P = (0, 0) Q = (0, 3) R = (4, 0)

¦PQ¦ = 3 ¦QR¦ = 5 ¦RP¦ = 4

5 (0, 0) + 4 (0, 3) + 3 (4, 0) (12, 12 )
= (1, 1)
I= =
5+4+3 12

In order to find the radius, firstly we must obtain the segment IP:

QP RP + RP PQ
IP =
QR + RP + PQ

The radius of the inscribed circle is the distance from I to the side PQ:

IP § PQ RP § PQ
d (I , PQ ) = =
PQ + QR + RP
PQ

whence the ratio of radius follows:

radius of circumscribed circle 1 PQ QR RP
=
2 PQ + QR + RP
radius of inscribed circle


Altitudes and orthocentre

The altitude of a side is the segment perpendicular to this side (also called base)
which passes through the opposite vertex. The three altitudes of a triangle intersect on a
unique point called the orthocentre. Let us prove this statement calculating the
RAMON GONZALEZ CALVET
74

intersection H of two altitudes. Since H belongs to the altitude passing through the vertex
P and perpendicular to the base QR (figure 8.5), its equation is:

H = P + z QR z imaginary
Figure 8.5
because the product by an imaginary
number turns the vector QR over π/2, that
is, z PQ has the direction of the altitude. H
also belongs to the altitude passing through
Q and perpendicular to the base RP. Then
its equation is:

H = Q + t RP t imaginary

By equating both expressions:

P + z QR = Q + t RP

we arrive at a vector written as a linear combination of two vectors but with imaginary
coefficients:

z QR ’ t RP = PQ

In this equation we must resolve PQ into components with directions perpendicular to the
vectors QR and RP. The algebraic resolution follows the same way as for the case of real
linear combination. Let us multiply on the right by RP:

z QR RP ’ t RP2 = PQ RP

and on the left:

RP z QR ’ RP t RP = RP PQ

The imaginary numbers anticommute with vectors:
2
’ z RP QR + t RP = RP PQ

By adding both equalities we arrive at:

z ( QR RP ’ RP QR ) = PQ RP + RP PQ

PQ RP + RP PQ PQ · RP
z= =
QR RP ’ RP QR QR § RP

Analogously one finds t:

PQ · QR
t=
QR § RP
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 75

Both expressions differ from the coefficients of real linear combination in the fact
that the numerator is an inner product. The substitution in any of the first equations gives:

H = P + z QR = P + ( PQ RP + RP PQ ) ( QR RP ’ RP QR ) ’1 QR =

The vector QR anticommutes with the outer product, an imaginary number:

H = P ’ ( PQ RP + RP PQ ) QR ( QR RP ’ RP QR ) ’1 =

Extracting the area as common factor, we obtain:

H = [ P ( QR RP ’ RP QR ) ’ ( PQ RP + RP PQ ) QR ] ( QR RP ’ RP QR ) ’1

By using the fact that PQ = Q ’ P, etc., we arrive at:

H = ( P QR RP ’ Q RP QR ’ P PQ QR + R PQ QR ) ( QR RP ’ RP QR ) ’1

= ( P2 QR + Q2 RP + R2 PQ + P QR P + Q RP Q + R PQ R ) ( QR RP ’ RP QR ) ’1

= ( P P · QR + Q Q · RP + R R · PQ ) ( QR § RP ) ’1

This formula is invariant under cyclic permutation of the vertices. Therefore all the
altitudes intersect on a unique point,
the orthocentre.
The equation of the orthocentre Figure 8.6
resembles that of the circumcentre. In
order to see the relationship between
both, let us draw a line passing through
P and parallel to the opposite side QR,
another one passing through Q and
parallel to RP, and a third line passing
through R and parallel to PQ (figure
8.6). Let A be the intersection of the
line passing through Q and that passing
through R, B be the intersection of the
lines passing through R and P respectively, and C be the intersection of the lines passing
through P and Q respectively. The triangle ABC is directly similar to the triangle PQR
with ratio ’2:
1 1 1
PQ = ’ AB QR = ’ BC RP = ’ CA
2 2 2

and P, Q and R are the midpoints of the sides of the triangle ABC:

B+C C+A A+ B
P= Q= R=
2 2 2

Therefore the altitudes of the triangle PQR are the perpendicular bisectors of the
sides of the triangle ABC, and the orthocentre of the triangle PQR is the circumcentre of
RAMON GONZALEZ CALVET
76

ABC. In order to prove with algebra this obvious geometric fact, we must only deduce
from the former relations the following equalities and substitute them into the orthocentre
equation:
B3 ’ B C 2 + C B2 ’ C 3
P P · QR =
8
C 3 ’ C A2 + A C 2 ’ A3
Q Q · RP =
8
A ’ A B + B A2 ’ B 3
3 2
R R · PQ =
8

By adding the three terms, the cubic powers vanish. On the other hand, the area of
the triangle PQR is four times smaller than the area of the triangle ABC:

BC § CA
QR § RP =
4

H = ( ’B C2 + C B2 ’ C A2 + A C2 ’ A B2 + B A2 ) ( 2 BC § CA ) ’1 =

= ’ ( A2 BC + B2 CA + C2 AB ) ( 2 BC § CA ) ’1

This is just the equation of the circumcentre of the triangle ABC.


Euler's line

The centroid G, the circumcentre O and the orthocentre H of a triangle are always
aligned on the Euler's line. To prove this fact, observe that the circumcentre and
orthocentre equations have the triangle area as a "denominator", while the centroid
equation does not, but we can introduce it:

P+Q + R
= ( P + Q + R ) QR § RP ( 3 QR § RP ) ’1
G=
3

Introducing the equality:

P Q ’Q P +Q R ’ R Q + R P ’ P R
QR § RP = P § Q + Q § R + R § P =
2

the centroid becomes:

G = ( P + Q + R ) ( P Q ’ Q P + Q R ’ R Q + R P ’ P R ) ( 6 QR § RP ) ’1

=(PPQ’PQP+PRP’PPR+QPQ’QQP+QQR’QRQ+

R Q R ’ R R Q + R R P ’ R P R ) ( 6 QR § RP ) ’1

= ( ’ P2 QR + P QR P ’ Q2 RP + Q RP Q ’ R2 PQ + R PQ R ) ( 6 QR § RP ) ’1
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 77

from where the relation between the three points G, H and O follows immediately:

H +2O
G=
3

Hence the centroid is located between the orthocentre and the circumcentre, and its
distance from the orthocentre is double of its distance from the circumcentre.


The Fermat's theorem

The geometric algebra allows to prove through a very easy and intuitive way the
Fermat's theorem.
Over every side of a triangle ABC we draw an equilateral triangle (figure 8.7). Let
T, U and S be the vertices of the
Figure 8.7
equilateral triangles respectively
opposite to A, B and C. Then the
segments AT, BU and CS have the same
length, form angles of 2π/3 and intersect
on a unique point F, called the Fermat's
point. Moreover, the addition of the
three distances from any point P to each
vertex is minimal when P is the Fermat's
point, provided that any of the interior
angles of the triangle ABC be higher
than 2π/3.
Firstly, we must demonstrate that
BU is obtained from AT by means of a rotation of 2π/3, which will be represented by the
complex number t:

2π 2π
t = cos + e12 sin
3 3

AT t = (AC + CT ) t = AC t + CT t

By construction, the vector AC turned over 2π/3 is the vector CU, and CT turned over
2π/3 is BC, so:

AT t = CU + BC = BU

Analogously, one finds CB = BU t and AT = CS t . That is, the vectors CS, BU and AT
have the same length and each one is obtained from each other by successive rotations of
2π/3.
Let us see that the sum of the distances from P to the three vertices A, B and C is
minimal when P is the Fermat™s point. We must prove firstly that the vectorial sum of PA
turned 4π/3, PB turned 2π/3 and PC is constant independently of the point P. (figure 8.8).
That is, for any two points P and P™ it is always true that:
RAMON GONZALEZ CALVET
78

PA t2 + PB t + PC = P'A t2 + P'B t + P'C

fact which is proved by arranging all the terms in one side of the equation:

PP' ( t2 + t + 1 ) = 0

This product is always zero since t2 + t + 1 = 0. Hence, there is a unique point Q such
that:

PA t2 + PB t + PC = QC
Figure 8.8
For any point P, the three segments form a
broken line as that shown in figure 8.8.
Therefore, by the triangular inequality we
have:

¦PA¦ + ¦PB¦ + ¦PC¦ ≥ ¦QC¦

When P is the Fermat™s point F, these
segments form a straight line. Then, the
addition of the distances from F to the three
vertices is minimal provided that no angle of
the triangle be higher than 2π/3:

¦FA¦ + ¦FB¦ + ¦FC¦ = ¦QC¦ ¤ ¦PA¦ + ¦PB¦ + ¦PC¦

In the other case, someone of the vectors FA t2, FB t or FC has a sense opposite to the
others, so that its length is subtracted from the others and their sum is not minimal.


Exercises

8.1 The Napoleon™s theorem. Over each side of a generic triangle draw an equilateral
triangle. Prove that the centres of the three equilateral triangles also form an
equilateral triangle.

8.2 Prove the Leibniz™s theorem. Let P be any point on the plane and G the centroid of a
triangle ABC. Then the following equality holds:

1
3 PG2 = PA2 + PB2 + PC2 ’ ( AB2 + BC2 + CA2 )
3

8.3 Let A, B and C be three given points on the plane. Then any point G on the plane can
be expressed as a linear combination of these three points (G is also considered as the
centre of masses located at A, B and C with weights a, b and c 4 ).


4
See August Ferdinand Möbius, Der Barycentrische Calcul, Leipzig, 1827, p. 17 (facsimile
edition of Georg Olms Verlag, Hildesheim,1976).
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 79

G=aA+bB+cC with a + b + c = 1

Prove that:
a) a, b, c are the fractions of the area of the triangle ABC occupied by the
triangles GBC, GCA and GAB respectively.
b) the geometric locus of the points P on the plane such that:

a PA2 + b PB2 + c PC2 = k

is a circle with centre G (Apollonius™ lost theorem).

8.4 Over every side of a convex quadrilateral ABCD, we draw the equilateral triangles
ABP, BCQ, CDR and DAS. Prove that:
a) If ¦AC¦ = ¦BD¦ then PR is perpendicular to QS.
b) If ¦PR¦ = ¦QS¦ then AC is perpendicular to BD.

8.5 Show that the length of a median of a triangle ABC passing through A can be
calculated from the sides as:

AB 2 + AC 2 BC 2
m= ’
2
A
2 4

8.6 In a triangle ABC we draw the bisectors of the angles A and B. Through the vertex C
we draw the lines parallel to each angle bisector. Let us denote the intersections of
each parallel line with the other bisector by D and E. Prove that if the line DE is
parallel to the side AB then the triangle ABC is isosceles.

8.7 The bisection of a triangle (proposed by Cristóbal Sánchez Rubio). Let P and Q be
two points on different sides of a triangle such that the segment PQ divides the
triangle in two parts with the same area. Calculate the segment PQ in the following
cases:
a) PQ is perpendicular to a given direction.
b) PQ has minimum length.
c) PQ passes through a given point inside the triangle.
RAMON GONZALEZ CALVET
80

9. CIRCLES

Algebraic and Cartesian equations

A circle with radius r and centre F is the locus of the points located at a distance r
from F, so its equation is:

FP2 = r2
d(F, P) =¦FP¦= r ”

which can be written as:

(P ’ F)2 = r2 P2 ’ 2 P · F + F2 ’ r2 = 0


By introducing the coordinates of P = (x, y) and F = (a, b) one obtains the analytic
equation of the circle:

x2 + y2 ’ 2 a x ’ 2 b y + a2 + b2 ’ r2 = 0

For example, the circle whose equation is:

x2 + y2 + 4 x ’ 6 y + 9 = 0

has the centre F = (’2, 3) and radius 2.
There always exists a unique circle passing through any three non aligned points.
In order to obtain this circle, one may substitute the coordinates of the points on the
Cartesian equation of the circle, arriving at a system of three linear equations of three
unknown quantities a, b and c:

± x1 2 + y1 2 + a x1 + b y1 + c = 0
2 2
 x2 + y 2 + a x2 + b y 2 + c = 0
x 2 + y 2 + a x + b y + c = 0
3 3 3 3



This is the classic way to find the equation of the circle, and from here the centre of the
circle.
However, a more direct way is the calculation of the circumcentre of the triangle
whose vertices are the three given points (see the previous chapter). Then the radius is the
distance from the centre to any vertex.


Intersections of a line with a circle

The coordinates of the intersection of a line with a circle are usually calculated by
substitution of the line equation into the circle equation, which leads to a second degree
equation. Let us now follow an algebraic way. Let F be the centre of the circle with radius
r, and R and R™ the intersections of the line passing through two given points P and Q
(figure 9.1):
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 81


Figure 9.1
± FR 2 = r 2

 R = P + k PQ

From the first equality we obtain:

(R ’ F )2 = R2 ’ 2 R · F + F2 = r2

The substitution of R given by the second
equality yields:

( P + k PQ )2 ’ 2 ( P + k PQ ) · F + F2 = r2

P2 + 2 k P · PQ + k2 PQ2 ’ 2 P · F ’ 2 k PQ · F + F2 ’ r2 = 0

By arranging the powers of k one obtains:

k2 PQ2 + 2 k ( P · PQ ’ F · PQ ) + P2 ’ 2 F · P + F2 ’ r2 = 0

Taking into account that PF = F ’ P one finds:

k2 PQ2 ’ 2 k PF · PQ + PF2 ’ r2 = 0

This equation has solution whenever the discriminant be positive:

4 ( PF · PQ )2 ’ 4 PQ2 ( PF2 ’ r2 ) ≥ 0

Introducing the identity (PF · PQ )2 ’ (PF § PQ )2 = PQ2 PF2 the discriminant becomes:

4 (PF § PQ )2 + 4 PQ2 r2 ≥ 0

The solution of the second degree equation for k is:

k = PF · PQ ’1 ± r 2 PQ ’ 2 + (PF § PQ ’1 )
2




When r = ¦PF § PQ¦/¦PQ¦, the line is tangent to the circle. In this case the height of the
parallelogram formed by the vectors PF and PQ is equal to the radius of the circle.
If u denote the unitary direction vector of the line PQ (figure 9.1):

PQ
u=
PQ

then both intersection points are written as:
RAMON GONZALEZ CALVET
82


R = P + kPQ = P + u « PF · u ’ r 2 + (PF § u ) 
2
¬ ·
 
R' = P + k' PQ = P + u « PF · u + r 2 + (PF § u ) 
2
¬ ·
 


Power of a point with respect to a circle

Both intersection points R and R' have the following property: the product of the
vectors going from a given point P to the intersections R and R' of any line passing
through P with a given circle is constant:

Figure 9.2
PR PR' = ( PF · u )2 ’ r2 ’ (PF § u )2

= PF2 ’ r2 = PT2

where T is the contact point of the tangent
line passing through P (figure 9.2). The
product PR PR™, which is independent of
the direction u of the line and only
depends on the point P, is called the
power of the point P with respect to the
given circle. The power of a point may be
calculated through the substitution of its coordinates into the circle equation. If its centre
is F = (a, b) and P = (x, y), then:

FP2 ’ r2 = ( x ’ a )2 + ( y ’ b )2 ’ r2 = x2 + y2 ’ 2 a x ’ 2 b y + a2 + b2 ’ r2

The power is equal to zero when P belongs to the circle, positive when P lies outside the
circle, and negative when P lies inside the circle.


Polar equation

We wish to describe the distance
from a point P to the points R and R' on the
circle as a function of the angle ± between
the line PR and the diameter (figure 9.3).
The vectors PR and PR' are:

PR = u « PF · u ’ r 2 ’ PF § u 
2
¬ ·
 
PR' = u « PF · u + r 2 + PF § u 
2
¬ · Figure 9.3
 

We arrive at the polar equation by taking the modulus of PR and PR', which is the
distance to the circle:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 83


PR = PF cos ± ’ r 2 ’ PF 2 sin 2±
Figure 9.4

PR' = PF cos ± + r 2 ’ PF 2 sin 2±

If P is a point on the circle then ¦PF¦ = r
(figure 9.4) and the equation is simplified:

¦PR¦ = 2 r cos ±


Inversion with respect to a circle

The point P' is the inverse point of P
with respect to a circle of centre F and radius r if the vector FP™ is the inverse vector of
FP with radius k:

FP' = k2 FP ’1 FP FP' = k2


Obviously, the product of the
Figure 9.5
vectors FP and FP' is only real and
positive when the points P and P™ are
aligned with the centre F and are located
at the same side of F. The circle of centre
F and radius k is called the inversion
circle because its points remain invariant
under this inversion. The inversion
transforms points located inside the
inversion circle into outside points and
reciprocally.
To obtain geometrically the
inverse of an inside point P, draw firstly
(figure 9.5) the diameter passing through P; after this draw the perpendicular to this
diameter from P which will cut the circle in the point T; finally draw the tangent with
contact point T. The intersection P' of this tangent with the prolongation of the diameter is
the inverse point of P. Note that the right triangles FPT and FTP' are oppositely similar:

FP FT ’1 = FP' ’1 FT ’ FP' FP = FT2 = k2

To obtain the inverse of an outside point, make the same construction but in the
opposite sense. Draw the tangent to the circle passing through the point (P' in the figure
9.5) and then draw the perpendicular to the diameter FP' passing through the contact point
T of the tangent. The intersection P of the perpendicular with the diameter is the searched
inverse point of P'.
Every line not containing the centre of inversion is transformed into a circle
passing through the centre (figure 9.6) and reciprocally. To prove this statement let us take
RAMON GONZALEZ CALVET
84

the polar equation of a straight line:

d
FP = Figure 9.6
cos ±

Then the inverse of P with centre F has the
equation:

k 2 cos ±
k2
FP' = =
FP d

which is the polar equation of a circle passing
through F and with diameter k2 / d . ± is the angle
between FP and FD, the direction perpendicular to
the line; therefore, the centre O of the circle is located on FD.
Let us prove that an inversion transforms a circle not passing through its centre
into another circle also not passing through its centre. Consider a circle of centre F and
radius r which will be transformed under an inversion of centre P and radius k. Then any
point R on the circle is mapped into another point S according to:

k2 k2
PS = =
PR PF cos ± ± r 2 ’ PF 2 sin 2±

where the polar equation of the circle is used. The multiplication by the conjugate of the
denominator gives:

)
(
k 2 PF cos ± m r 2 ’ PF 2 sin 2±
PS =
PF 2 ’ r 2

which is the polar equation of a circle with centre G and radius s:

PS = PG cos ± m s 2 ’ PG 2 sin 2±

k 2 PF k 2 PF k2 r k2 r
PG = = s= =
where
PF 2 ’ r 2 PF 2 ’ r 2 PT 2
PT 2

PF2 ’ r2 = PT2 is the power of the point P with respect to the circle of centre F.
Then we see that the distance from the centre of inversion to the centre of the circle and
the radius of this circle changes in the ratio of the square of the radius of inversion divided
by the power of the centre of inversion with respect to the given circle.
What is the power of the centre of inversion P with respect to the new circle?:

k 4 (PF 2 ’ r 2 ) k4
PU = PG ’ s = =
2 2 2

PT 4 PT 2

That is, the product of the powers of the centre of inversion with respect to any circle and
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 85

its transformed is constant and equal to the fourth power of the radius of inversion:

PT2 PU2 = k4

In the equation of the transformed circle the symbol m appears instead of ± . It
means that the sense of the arc from R to R' is opposite to the sense of the arc from S to S'.


The nine-point circle

The Euler™s theorem states that the midpoints of the sides of any triangle, the feet
of the altitudes and the midpoints from each vertex to the orthocentre lie on a circle called
the nine-point circle.
Let any triangle PQR whose orthocentre is H be. Now we evaluate the expression:

(P ’ H + Q ’ R )2 ’ (P ’ H ’ Q + R )2 = 4(P ’ H ) · (Q ’ R ) = 0

which vanishes because the segment PH lying on the altitude passing through P is
perpendicular to the base QR. Analogously:

(P ’ H + Q ’ R )2 ’ (P + H ’ Q ’ R )2 = 4(Q ’ H ) · (P ’ R ) = 0
and
(P + H ’ Q ’ R )2 ’ (P ’ H ’ Q + R )2 = 4(R ’ H ) · (Q ’ P ) = 0

(P + Q ’ R ’ H )2 = (P ’ Q + R ’ H )2 = (P ’ Q ’ R + H )2
Hence:

Since the sign may be opposite inside the square we have:

(P + Q ’ R ’ H )2 = (P ’ Q + R ’ H )2 = (P ’ Q ’ R + H )2 =

= (’ P ’ Q + R + H ) = (’ P + Q ’ R + H ) = (’ P + Q + R ’ H )
2 2 2




Now we introduce the point N = (P + Q + R + H )/4 to obtain:
2 2 2 2 2 2
R+H Q+H Q+ R P+Q P+ R P+H
« « « « « «
¬N ’ · = ¬N ’ · = ¬N ’ · = ¬N ’ · = ¬N ’ · = ¬N ’ ·
2 2 2 2 2 2
     

That is, N is the centre of a circle (figure 9.7) passing through the three midpoints of the
sides (P + Q)/2, (Q + R)/2 and (R + P)/2 and the three midpoints of the vertices and the
orthocentre (P + H)/2, (Q + H)/2 and (R + H)/2.
Since we can write:
RAMON GONZALEZ CALVET
86

1 « P + H  1 «Q + R
N= ·+ ¬
¬ ·
2 2  2 2 

it follows that (P+H)/2 and (Q+R)/2 are
Figure 9.7
opposite extremes of the same diameter.
Then the vectors J’(P+H)/2 and
J’(Q+R)/2, which are orthogonal:

P+H Q + R
« «
¬J ’ ··¬J ’ ·=0
2 2
 

are the sides of a right angle which
intercepts a half circumference of the nine
point circle. Therefore, this angle is
inscribed and its vertex J also lies on the
nine-point circle. Also, we may go
through the algebraic way to arrive at the
same conclusion. Developing the former inner product:

Q+R P+H P+H Q+R
J2 ’J · ’ ·J + =0
·
2 2 2 2

Introducing the centre N of the nine-point circle, we have:
2
Q + R «Q + R
J ’2J ·N +2N · ’¬ · =0
2

2 2


and after adding and subtracting N2 we arrive at:
2
«Q + R 
(J ’ N ) 2
’¬ ’N· =0
2 

which shows that the length of JN is the radius of the nine-point circle.
From the formulas for the centroid and orthocentre one obtains the centre N of the
nine-point circle:

3G + H
= (P QR P + Q RP Q + R PQ R )(4 PQ § QR )
’1
N=
4
Figure 9.8
that is, the point N also lies on the Euler™s line. The
relative distances between the circumcentre O, the
centroid G, the nine-point circle centre N and the
orthocentre H are shown in the figure 9.8.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 87

Cyclic and circumscribed quadrilaterals

First, let us see a statement valid for every quadrilateral: the area of a quadrilateral
ABCD is the outer product of the diagonals AC and BD. The prove is written in one line.
Since we can divide the quadrilateral in two triangles we have:

1 1 Figure 9.9
AB § BC + AC § CD
Area =
2 2
1
= (AC § BC + AC § CD )
2
1
= AC § BD
2
1
= ¦AC¦¦BD¦sin(AC, BD) e12
2

The quadrilaterals inscribed in a circle are called cyclic quadrilaterals. They fulfil
the Ptolemy™s theorem: the product of the lengths of both diagonals is equal to the
addition of the products of the lengths of opposite sides (figure 9.9):

BD = AB CD + BC
AC DA

A beautiful demonstration makes use of the cross ratio and it is proposed in the first
exercise of the next chapter.
For the cyclic quadrilaterals, the sum of opposite angles is equal to π, because they
are inscribed in the circle and intercept opposite arcs. Another interesting property is the
Brahmagupta formula for the area. If the lengths of the sides are denoted by a, b, c and d
and the semiperimeter by s then:

a+b+c+d
(s ’ a ) (s ’ b) (s ’ c ) (s ’ d )
Area = s=
2

With this notation, the proof is written more briefly. The law of cosines applied to each
triangle of quadrilateral gives:

± AC 2 = a 2 + b 2 ’ 2 a b cos »

 AC = c + d ’ 2 c d cos( π ’ » )
2 2 2




Equating both equations taking into account that the cosines of supplementary angles have
opposite sign, we have:

a 2 + b 2 ’ 2 a b cos » = c 2 + d 2 + 2 c d cos »

a 2 + b 2 ’ c 2 ’ d 2 = 2 ( a b + c d ) cos »
RAMON GONZALEZ CALVET
88

a 2 + b2 ’ c 2 ’ d 2
= cos »
2(ab+cd )

The area of the quadrilateral is the sum of the areas of the triangles ABC and CDA:

ab cd
sin( π ’ » )
sin » +
Area =
2 2

Both sines are equal; then we have:

(a + b2 ’ c2 ’ d 2 ) 
( a b + c d )2 (a b + c d )2 « 2
2
¬ ·
sin 2 » = 1’
Area 2 =
4(ab+cd )
¬ ·
2
4 4  

)2 ’ ( a 2 + b 2 ’ c 2 ’ d 2 )2
4(ab+cd
= (s ’ a )(s ’ b )(s ’ c )(s ’ d )
=
16

Figure 9.10
Another type of quadrilaterals are those
where a circle can be inscribed, also called
circumscribed quadrilaterals. For these
quadrilaterals the bisectors of all the angles
meet in the centre I of the inscribed circle,
because it is equidistant from all the sides.
When tracing the radii going to the tangency
points, the quadrilateral is divided into pairs of
opposite triangles (figure 9.10), and hence one
deduces that the sums of the lengths of
opposite sides are equal:


¦AB¦+¦CD¦= x + y + z + t = y + z + t + x =¦BC¦+¦DA¦

If a quadrilateral has inscribed and circumscribed circle, then we may substitute this
condition (a + c = b + d) in the Brahmagupta formula to find:

Area = a b c d = AB BC CD DA

Since the sum of the angles of any quadrilateral is 2π and either of the four small
quadrilaterals of a circumscribed quadrilateral has two right angles (figure 9.10), it
follows that the central angles are supplementary of the vertices:

CBA ≡ UBT = π ’ TIU ADC ≡ WDV = π ’ VIW

If the quadrilateral is also cyclic then the angles CBA and ADC are inscribed and intercept
opposite arcs of the outer circle, so that the segments joining opposite tangency points are
orthogonal:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 89

CBA + ADC = π ’ TIU + VIW = π ’ TV⊥UW


Angle between circles

Let us calculate the angle between a circle centred at O with radius r and another
one centred at O' with radius r'. If ¦O' ’ O¦< r + r' then they intersect. The points of
intersection P must fulfil the equations for both circles:

± (P ’ O )2 = r 2 ± P2 ’ 2 P ·O + O2 = r2
 2

(P ’ O' ) = r'
2
 P ’ 2 P · O' + O' = r'
2 2
2




Adding both equations we obtain:

2 P 2 ’ 2 P · (O + O' ) + O 2 + O' 2 = r 2 + r' 2

It is trivial that the angle of intersection is the supplementary of the angle between both
radius, its cosines being equal:

P 2 ’ P · (O + O' ) + O · O'
OP · O'P
cos ± = =
OP OP' r r'

The substitution of the first equality into the second gives:

r 2 + r' 2 ’ (O ’ O' )

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