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PR · n
1 1
= ¦’ n ’1 PR n ’ PR ¦= ¦’ n ’1 ( PR n + n PR ) ¦=
2 2 n

This expression is the modulus of the projection of the vector PR upon the perpendicular
direction. Both formulas of the distance are equivalent because the normal and direction
vectors anticommute:

nv=’vn ” ”
nv+vn=0 n·v=0

Written with components:

n1 v1 + n2 v2 = 0

From each vector the other one is easily obtained by exchanging components and altering
a sign:

n = ( n1 , n2 ) = ( v2 , ’v1 )

As an application, let us calculate the equation of the line r passing through the
points (2, 3) and (7, 6). A direction vector is the segment having as extremes both points:

v = (7, 6) ’ (2, 3) = (5, 3)

and an equation for this line is:

x ’2 y ’3
=
5 3

From the direction vector we calculate the perpendicular vector n and the general equation
of the line:

n = (3, ’5) 3 (x ’ 2) ’ 5 (y ’ 3) = 0 ” 3x’5y+9=0

Now, let us calculate the distance from the point R = (’2, 1) to the line r. We choose the
point (2, 3) as the point P on the line:

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

’ 4 … 3 + ( ’2) … ( ’5)
PR § v PR · n 2
=
d( (’2, 1), r ) = = =
v n 17
34

For another point R = (’3, 0 ), we will find a null distance indicating that this point lies on
r.
RAMON GONZALEZ CALVET
40


Slope and intercept equations of a line

If the ordinate is written as a function of the abscissa, the equation so obtained is
the slope-intercept equation:

y=mx+b

Note that this expression cannot describe the vertical lines whose equation is x = constant.
The coefficient m is called the slope because it is a measure of the inclination of the line.
For any two points P and Q on the line we have:

yP = m xP + b

yQ = m xQ + b

Subtracting both equations:

yQ ’ yP = m (xQ ’ xP)

we see that the slope is the quotient of
ordinate increment divided by the abscissa
increment (figure 5.9):
Figure 5.9
yQ ’ y P
m=
xQ ’ x P

This quotient is also the trigonometric tangent of the oriented angle between the line and
the positive abscissa semiaxis. For angles larger than π/2 the slope becomes negative.
b is the ordinate intercepted at the origin, the y-intercept:


x=0 y=b

If we know the slope and a point on a line, the point-slope equation may be used:

y ’ yP = m (x ’ xP)

Also, one may write for the intercept equation of a line:

xy
+ =1
ab

where a and b are the x-intercept and y-intercept of the line with each coordinate axis.


Polar equation of a line

In this equation, the distance from a fixed point F to the generic point P on the line
is a function of the angle ± with regard to the perpendicular direction (figure 5.10). If d is
the distance from F to the line, then:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 41



d
FP =
cos ± Figure 5.10

This equation allows to relate easily the
straight line with the circle and other conic
sections.


Intersection of two lines and pencil of
lines

The calculation of the intersection
of two lines is a very usual problem. Let us
suppose that the first line is given by the point P and the direction vector u and the second
one by the point Q and the direction vector v. Denoting by R the intersection point, which
belongs to both lines, we have:

” k u ’ l v = Q ’ P = PQ
R=P+ku=Q+lv

In order to find the coefficients k and l, the vector PQ must be resolved into a linear
combination of u and v, what results in:

k = PQ § v (u § v) ’1 l = ’ u § PQ (u § v) ’1

The intersection point is also obtained by directly solving the system of the
general equations of both lines:

± n1 x + n 2 y + c = 0

n1' x + n 2' y + c' = 0

The pencil of lines passing through this point is the set of lines whose equations
are linear combinations of the equations of both given lines:

(1 ’ p) [ n1 x + n2 y + c ] + p [ n'1 x + n'2 y + c' ] = 0

where p is a real parameter and
’ ∞ ¤ p ¤ ∞ . Each line of the pencil
determines a unique value of p,
independently of the fact that the
general equation for a line is not
unique
If the equation system has a
unique point as solution, all the lines
of the pencil are concurrent at this
point. But in the case of an
incompatible system, all the lines of Figure 5.11
the pencil are parallel, that is, they
RAMON GONZALEZ CALVET
42


intersect in a point at infinity, which gives generality to the concept of pencil of lines.
If the common point R is known or given, then the equation of the pencil is written
as:

[ ( 1 ’ p ) n1 + p n'1 ] ( x ’ xR ) + [ ( 1 ’ p ) n2 + p n'2 ] ( y ’ yR ) = 0

For 0<p<1, the normal and direction vectors of the p-line are respectively comprised
between the normal and direction vectors of both given lines4. In the other case the vectors
are out of this region (figure 5.11)
For example, calculate the equation of the line passing through the point (3, 4) and
the intersection of the lines 3 x + 2 y + 4 = 0 and 2 x ’ y + 3 = 0. The equation of the
pencil of lines is:

(1’p)(3x+2y+4)+p(2x’y+3)=0

The line of this pencil to which the point (3, 4) belongs must fulfil:

(1’p)(3…3+2…4+4)+p(2…3’4+3)=0

yielding p = 21/16 and the line:

27 x ’ 31 y + 43 = 0

Which is the meaning of the coefficient of linear combination p? Let us write the
pencil of lines P determined by the lines R and S as:

P=(1’p)R+pS

which implies the same relation for the normal vectors:

nP = ( 1 ’ p) nR + p nS

Taking outer products we obtain:

nP § nR nP § nR
p
= p=

nP § ( nR ’ nS )
nP § nS p ’ 1

These equalities are also valid for direction vectors whenever they have the same modulus
than the normal vectors. When the modulus of the normal vectors of the lines R and S are
equal we can simplify:

sin (RP )
p
=
1 ’ p sin (PS )

In this case, the value p = 1/2 corresponds to the bisector line of R and S. We see from this
expression and the foregoing ones that each value of p corresponds to a unique line.
4
This statement requires that the angle from the direction vector to the normal vector be a positive
right angle.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 43



Dual coordinates

The duality principle states that any theorem relating incidence of lines and points
implies a dual counterpart where points and lines have been exchanged. The phrase «two
points determine a unique line passing through these points» has the dual statement «two
lines determine a unique point, intersection of these lines». All the geometric facts have an
algebraic counterpart, and the duality is not an exception. With the barycentric coordinates
every point R on the plane can be written as linear combination of three non aligned points
O, P and Q:

R = ( 1 ’ p ’ q ) O + p P + q Q = (p, q)

where p and q are the coordinates and O the origin.
These three points determine three non parallel lines A, B, and C in the following
way:
A = PQ B = QO C = OP

Then the direction vectors vA = PQ, vB = OP, vC = QO are related by:

v A + v B + vC = 0

Note that vB and ’vC are the base of the vectorial plane.
Any line D on the plane can be written as linear combination of these lines:

D=(1’b’c)A+bB+cC

This means that the general equation (with point coordinates) of D is a linear combination
of the general equations of the lines A, B and C. I call b and c the dual coordinates of the
line D. In order to distinguish them from point coordinates, I shall write D = [b, c]. The
choice of the equation for each line must be unambiguous and therefore the normal vector
in the implicit equations for A, B and C will be obtained by turning the direction vector
over π/2 counterclockwise.
Let us see some special cases. If the dual coordinate b is zero we have:

D=(1’c)A+cC

which is the equation of the pencil of lines passing through the intersection of the lines A
and C, that is, the point P. Then P = [0, c] (for every c). Analogously, c = 0 determines the
pencil of lines passing through the intersection of the lines A and B, which is the point Q,
and then Q = [b, 0] (for every b). The origin of coordinates is the intersection of the lines
B and C and then O = [b, 1 ’ b] (for every b). Compare the dual coordinates of these
points:

O = [b, 1’b] ∀ b, c
P = [0, c] Q = [b, 0]

with the point coordinates of the lines A, B and C:
RAMON GONZALEZ CALVET
44


A = (p, 1’ p) ∀ p, q
B = (0, q) C = (p, 0)

Let us see an example: calculate the dual Cartesian coordinates of the line 2x + 3y
+ 4 = 0. The points of the Cartesian base are O = (0, 0), P = (1, 0) and Q = (0, 1). Then the
lines of the Cartesian base are A: ’x ’ y +1= 0, B: x = 0, C: y = 0. We must solve the
identity:

∀ x, y
2 x + 3 y + 4 ≡ a' ( ’ x ’ y + 1) + b' x + c' y

x ( 2 + a' ’ b' ) + y ( 3 + a' ’ c' ) + 4 ’ a' ≡ 0

whose solution is:

a' = 4 b' = 6 c' = 7

Dividing by the sum of the coefficients we obtain:

2x +3y +4 4 6 7
≡ ( ’ x ’ y +1)+ x+ y
17 17 17 17

from where the dual coordinates of this line are obtained as [b, c] = [6/17, 7/17]. Let us
see their meaning. The linear combination of both coordinates axes is a line of the pencil
of lines passing through the origin:

6 7
x+ y=0 6x+7y=0
or
13 13

This line intersects the third base line ’x ’y + 1 = 0 at the point (7, ’6), whose pencil of
lines is described by:

«6 7
a ( ’ x ’ y + 1) + ( 1 ’ a ) ¬ x + y· = 0
 13 13 

Then 2 x + 3 y + 4 = 0 is the line of this pencil determined by a = 4/17.
On the other hand, how may we know whether three lines are concurrent and
belong to the same pencil or not? The answer is that the determinant of the dual
coordinates must be zero:

D=(1’b’c)A+bB+cC

E = ( 1 ’ b' ’ c' ) A + b' B + c' C

F = ( 1 ’ b'' ’ c'' ) A + b'' B + c'' C
1’ b ’ c b c
1 ’ b' ’ c' c' = 0
b'

D, E and F concurrent
1 ’ b'' ’ c'' b'' c''
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 45


When it happens, we will say that the lines are linearly dependent, and we can write
anyone of them as linear combination of the others:

F=(1’k)D+kE

We can express a point with an equation for dual coordinates in the same manner
as we express a line with an equation for point coordinates. For example, the point (5, 3)
is the intersection of the lines x = 5, y = 3 whose dual coordinates we calculate now:

x ’ 5 ≡ a' ( 1 ’ x ’ y ) + b' x + c' y ’ a' = ’5 b' = ’4 c' = ’5

From where a = 5/14, b = 4/14 and c = 5/14. The dual coordinates of x = 5 are [4/14,
5/14]. Analogously:

y ’ 3 ≡ a' ( 1 ’ x ’ y ) + b' x + c' y ’ a' = ’3 b' = ’3 c' = ’2

From where a = b = 3/8, c = 2/8. The dual coordinates of y = 3 are [3/8, 2/8]. The linear
combinations of both lines are the pencil of the point (5, 3), which is described by the
parametric equation:

[b, c] = ( 1 ’ k ) ® 4 ,
5 ®3 2
+k  , 
°14 14 
 °8 8 »
»

By removing the parameter k, the general equation is obtained:

b ’ 4 / 14 c ’ 5 / 14
= 12 b + 10 c ’ 7 = 0

’6
5

That is, the dual direction vector of the point (5, 3) is [5, ’6] and the dual normal vector is
[6, 5]. In the dual plane, an algebra of dual vectors can be defined. A dual direction vector
for a point may be obtained as the difference between the dual coordinates of two lines
whose intersection be the point. Then, there exist dual translations of lines:

dual + : L — W ’ L L ={plane lines} W = {dual vectors}

(A, w) ’ B = A + w

That is, we add a fixed dual vector w to the line A in order to obtain another line B.
Let us prove the following theorem: all the points whose dual direction vectors are
proportional are aligned with the centroid of the coordinate system .
The proof begins from the dual continuous equation for a point P:

b ’ b0 c ’ c 0
=
v1 v2

which can be written in a parametric form:
RAMON GONZALEZ CALVET
46


[b, c] = (1 ’ k )[b0 , c0 ] + k [b0 +`v1 , c 0 + v 2 ]
P:

This equation means that P is the intersection point of the lines with dual coordinates [b0,
c0] and [b0 +v1, c0 +v2]. Then P must be obtained by solving the system of equations of
each line for the point coordinates p, q (x, y if Cartesian):

(1 ’ b0 ’ c 0 ) A + b0 B + c 0 C = 0
±

(1 ’ b0 ’ c 0 ’ v1 ’ v 2 ) A + (b0 + v1 ) B + (c 0 + v 2 ) C = 0

By subtraction of both equations, we find an equivalent system:

±(1 ’ b0 ’ c 0 ) A + b0 B + c 0 C = 0

 ’ (v1 + v 2 ) A + v1 B + v 2 C = 0

Now, if we consider a set of points with the same (or proportional) dual direction vector,
the first equation changes but the second equation remains constant (or proportional). That
is, the first line changes but the second line remains constant. Therefore all points will be
aligned and lying on the second line:

’ (v1 + v 2 ) A + v1 B + v 2 C = 0

However in which matter do two points differ whether having or not a proportional dual
direction vector? We cannot say that two points are aligned, because we need three points
at least. Let us search the third point X, rewriting the foregoing equality:

v1 v
(B ’ A) + 2 (C ’ A) = 0
v1 + v 2 v1 + v 2

Now the variation of the components of the dual vector generates the pencil of the lines B
’ A and C ’ A. That is, the intersection of both lines is the point X searched:

±B ’ A = 0
X: 
C ’ A = 0

Then, two points have a proportional dual direction vector if they are aligned with the
point X, intersection of the lines B ’ A and C ’ A. However, note that the addition of
coefficients of each line is zero instead of one, so that one dual coordinate of each line is
infinite:

B ’ A = [∞, c ] C ’ A = [b, ∞]

I call X the point at the dual infinity or simply the dual infinity point. The dual infinity
point has finite coordinates and a very well defined position. Then the points with
proportional dual vector are always aligned with the dual infinity point, and I shall say
that they are parallel points in a dual sense, of course. In order to precise which point is X
let us take in mind that the lines B ’ A and C ’ A are the medians of the triangle OPQ (the
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 47


point base). Hence the dual infinity point is the centroid of the base triangle OPQ. With
Cartesian coordinates O = (0, 0), P = (1, 0), Q = (0, 1) and X = (1/3, 1/3). This ends the
proof.
Summarising, we can say that two parallel points are aligned with the dual infinity
point, which is the dual statement of the fact that two parallel lines meet at the infinity,
that is, there is a line at the infinity, in the usual sense. This is the novelty of the projective
geometry in comparison with the Euclidean geometry. In fact, the problem arises because
the point coordinates take infinite values, but the line L at the infinity is a well defined line
with finite dual coordinates [1/3, 1/3]:

®1 1 A + B + C
L= , =
° 3 3» 3

This means that the line L at the infinity belongs to the pencil of the lines (A + B ) /2 and C
. But both lines are parallel because their direction vectors are proportional:

v A + v B = ’v C

That is, the lines (A + B ) /2 and C meet at a point located at an infinite distance from the
origin of coordinates. Since any other parallel line meets them also at the infinity, this
argument does not suffice. However we can understand that the line L also belongs to the
pencil of the lines A and (B + C ) /2, which are also parallel with another direction, that is,
they meet at another point of the infinity. Any other pencil we take has always its point of
intersection located at the infinity. Then L only has points located at the infinity and
because of this it is called the line at the infinity. Summarising we can say that the line at
the infinity is the centroid of the three base lines in spite of the incompatibility of its
equation:

(’ x ’ y + 1) + x + y = 0
A+ B +C 1
L= =0
’ ’
3 3 3

Moreover, we may interpret the parallel lines as those lines aligned (in a dual sense) with
the line at the infinity:

E = [bE, cE] F = [bF, cF]

bF ’ b E b F ’ 1 / 3
=

E || F
cF ’ cE cF ’ 1 / 3

This is a useful equality because it allows us to know whether two lines are parallel from
their dual coordinates.
Many of the incidence theorems (and also their dual theorems) can be solved by
means of line equations or dual coordinates. A proper example is the proof of the
Desargues theorem.


The Desargues theorem
RAMON GONZALEZ CALVET
48


Given two triangles ABC and A'B'C', let P be the intersection of the prolongation
of the side AB with the side A'B', Q the intersection of BC with B'C', and R the intersection
of CA with C'A'. The points P, Q and R are aligned if and only if the lines AA', BB' and
CC' meet at the same point.

Proof ’ The hypothesis states that the lines AA', BB' and CC' intersect at the same
point O (figure 5.12):

O = a A + ( 1 ’ a ) A'

O = b B + ( 1 ’ b ) B'

O = c C + ( 1 ’ c ) C'

with a, b and c being real. Equating
the first and second equations we
obtain:
Figure 5 12
a A + ( 1 ’ a ) A' = b B + ( 1 ’ b ) B'

which can be rearranged as:

a A ’ b B = ’ ( 1 ’ a ) A' + ( 1 ’ b ) B'

Dividing by a ’ b the sum of the coefficients becomes the unity, and then the equation
represents the intersection of the lines AB and A'B', which is the point P:

a ’1 b ’1
a b
P= A’ B= A' ’ B'
a’b a’b a’b a’b

By equating the second and third equations for the point O, and the third and first ones,
analogous equations for Q and R are obtained:

b ’1 c ’1
b c
Q= B’ C= B' ’ C'
b’c b’c b’c b’c

c ’1 a ’1
c a
R= C’ A= C' ’ A'
c’a c’a c’a c’a

Now we must prove that P, Q and R are aligned, that is, fulfil the equation:

R=dP+(1’d)Q with d real

With the substitution of the former equations into the last one we have:

®a  ®b 
c a b c
B + ( 1 ’ d ) 
C’ A=d  A’ B’ C
c’a c’a °a ’ b b’a » °b ’ c b’c »
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 49


Arranging all the terms at the left hand side, the expression obtained must be identical to
zero because A, B and C are non aligned points:

b (1 ’ d )  c (1 ’ d ) 
® da a ® db ®c
’ ’ A+  ’ B+ + C ≡0
 a ’ b c ’ a b’c  b’c 
a’b c’a
° » ° » ° »

This implies that the three coefficients must be null simultaneously yielding a unique
value for d,

a’b
d=
a’c

fact which proves the alignment of P, Q and R, and gives the relation for the distances
between points:

QR QP ’1 = ( A'O A'A ’1 ’ B'O B'B ’1 ) ( A'O A'A-1 ’ C'O C'C ’1 ) ’1

Proof ⇐ We will prove the Desargues theorem in the other direction following the
same algebraic way but applying the duality, that is, I shall only change the words. O, A,
B, C, P, Q and R will be now lines (figure 5.13).
The hypothesis states that the
points AA', BB' and CC' belong to the
same line O, that is, O belongs to the
pencil of the lines A and A', but also
to the pencil of the lines BB' and CC':

O = a A + ( 1 ’ a ) A'

O = b B + ( 1 ’ b ) B'
Figure 5.13
O = c C + ( 1 ’ c ) C'

with a, b and c being real
coefficients. Equating the first and
second equations we obtain:

a A + ( 1 ’ a ) A' = b B + ( 1 ’ b ) B'

which can be rearranged as:

a A ’ b B = ’ ( 1 ’ a ) A' + ( 1 ’ b ) B'

Dividing by a ’ b the sum of the coefficients becomes the unity, and then the equation
represents the line passing through the points AB and A'B' (belonging to both pencils),
which is the line P:

a ’1 b ’1
a b
P= A’ = A' ’ B'
a’b a’b a’b a’b
RAMON GONZALEZ CALVET
50



By equating the second and third equations for the line O, and the third and first ones,
analogous equations for Q and R are obtained:

b ’1 c ’1
b c
Q= B’ C= B' ’ C'
b’c b’c b’c b’c

c ’1 a ’1
c a
R= C’ A= C' ’ A'
c’a c’a c’a c’a

Now we must prove that the lines P, Q and R belong to the same pencil, that is, fulfil the
equation:

R=dP+(1’d)Q with d real

With the substitution of the former equations into the last one we have:

®a  ®b 
c a b c
B + ( 1 ’ d ) 
C’ A=d  A’ B’ C
c’a c’a °a ’ b b’a » °b ’ c b’c »

Arranging all the terms at the left hand side, the expression obtained must be identical to
zero because A, B and C are independent lines (not belonging to the same pencil):

b (1 ’ d )  c (1 ’ d ) 
® da a ® db ®c
’ ’ A+  ’ B+ + C ≡0
 a ’ b c ’ a b’c  b’c 
°a ’ b °c ’ a
° » » »

This implies that the three coefficients must be null simultaneously yielding a unique
value for d,

a’b
d=
a’c

fact which proves that P, Q and R are lines of the same pencil.

Exercises

Let A = (2, 4), B = (4, ’3) and C = (2, ’5) be three consecutive vertices of a
5.1
parallelogram. Calculate the fourth vertex D and the area of the parallelogram.

Prove the Euler™s theorem: for any four points A, B, C and D the product AD BC +
5.2
BD CA + CD AB vanishes if and only if A, B and C are aligned.

Consider a coordinate system with vectors { e1, e2 } where ¦e1¦= 1, ¦e2¦= 1 and
5.3
the angle formed by both vectors is π/3.
a) Calculate the area of the triangle ABC being A = (2, 2), B = (4, 4), C = (4, 2).
b) Calculate the distance between A and B, B and C, C and A.

Construct a trapezoid whose sides ¦AB¦, ¦BC¦, ¦CD¦ and ¦DA¦ are known and
5.4
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 51


AB is parallel to CD. Study for which values of the sides does the trapezoid exist.

Given any coordinate system with points {O, P, Q} and any point R with
5.5
coordinates (p, q) in this system, show that:

Area OPR
Area RPQ Area ORQ
1’ p ’ q = p= q=
Area OPQ Area OPQ Area OPQ

Let the points A = (2, 3), B = (5, 4), C = (1, 6) be given. Calculate the distance
5.6
from C to the line AB and the angle between the lines AB and AC.

Given any barycentric coordinate system defined by the points {O, P, Q} and the
5.7
non aligned transformed points {O', P', Q'}, an affinity (or an affine
transformation) is defined as that geometric transformation which maps each point
D to D' in the following way:

D = ( 1 ’ x ’ y ) O + x P + y Q’ D' = ( 1 ’ x ’ y ) O' + x P' + y Q'

Prove that:
a) An affinity maps lines onto lines.
b) An affinity is equivalent to a linear mapping of the coordinates, that is, the
coordinates of any transformed point are linear functions of the coordinates of the
original point.
c) An affinity preserves the coordinates expressed in any other set of independent
points different of the given base:

D = ( 1 ’ b ’ c ) A + b B + c C’ D' = ( 1 ’ b ’ c ) A' + b B' + c C'
d) An affinity maps a parallelogram to another parallelogram, and hence, parallel
lines to parallel lines.
e) An affinity preserves the ratio DE DF ’1 for any three aligned points D, E and
F.

If {A, B, C} is a base of lines and {A', B', C'} their transformed lines -the lines of
5.8
each set being independent-, consider the geometric transformation which maps
each line D to D' in the following way:

D = ( 1 ’ b ’ c ) A + b B + c C’ D' = ( 1 ’ b ’ c ) A' + b B' + c C'

a) Prove that every pencil of lines is mapped to another pencil of lines.
b) The dual coordinates of D' are linear functions of the dual coordinates of D.
c) This transformation preserves the coefficients which express a line as a linear
combination of any three non concurrent lines, that is, in the foregoing mapping
{A, B, C} have not to be necessarily the dual coordinate base and can be any other
set of independent lines.
d) Parallel points are mapped to parallel points.
e) For any three concurrent lines P, Q and R, the single ratio of the dual vectors
PQ PR’1 is preserved.
f) Using the formula of the cross ratio of a pencil of any four lines (ABCD) as a
RAMON GONZALEZ CALVET
52


function of their direction vectors5:

v A § vC v B § v D
( ABCD ) =
v A § v D v B § vC

show that it is preserved.

Calculate the dual coordinates of the lines x “ y + 1 = 0 and x “ y + 3 = 0 . See
5.9
that they are aligned in the dual plane with the line at the infinity, whose dual
coordinates are [1/3, 1/3], and therefore are parallel.

5.10 Calculate the dual equations of the points (2, 1) and (“3, “1) and their direction
vectors. See that they are parallel points. Hence prove that they are aligned with
the dual infinity point, the centroid of the coordinate system (1/3, 1/3).




5
This formula is deduced in the chapter devoted to the cross ratio.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 53

6. ANGLES AND ELEMENTAL TRIGONOMETRY

Here, the basic identities of the elemental trigonometry are deduced in close
connection with basic geometric facts, a very useful point of view for our pupils.

Sum of the angles of a polygon

Firstly let us see the special case
of a triangle. For any triangle ABC the
following identity holds:

CA AB ’1 AB BC ’1 BC CA ’1 = 1

Let ±, β and γ be the exterior angles
between the sides CA and AB, AB and
Figure 6.1
BC, BC and CA respectively (figure 6.1).
Applying the definition of geometric
quotient, the modulus of all the sides are
simplified and only the exponentials of the arguments remain:

exp(± e12 ) exp(β e12 ) exp(γ e12 ) = exp[ (± + β + γ ) e12 ] = 1

The three angles have the same orientation, which we suppose positive, and are
lesser than π. Hence, since the exponential is equal to the unity, the addition of the three
angles must be equal to 2π:

± + β + γ = 2π

The interior angles, those formed by AB and AC, BC and CA, CA and CB are
supplementary of ±, β, γ (figure 6.1). Therefore the sum of the angles of a triangle is equal
to π:

(π ’ ± ) + (π ’ β ) + (π ’ γ ) = π

This result is generalised to any polygon from the following identity:

AB BC ’1 BC ... YZ ’1 YZ ZA’1 ZA AB ’1 = 1

Let ±, β, γ ... ω be the exterior angles formed by the sides ZA and AB, AB and BC,
BC and CD, ..., YZ and ZA, respectively. After the simplification of the modulus of all
vectors, we have:

exp[ (± + β + γ + ... + ω ) e12 ] = 1

Let us suppose that the orientation defined by the vertices A, B, C ... Z is
counterclockwise, although the exterior angles be not necessarily all positive1. Translating

1
That is, it is not needed that the polygon be convex.
RAMON GONZALEZ CALVET
54

them to a common vertex, each angle is placed close each other following the order of the
perimeter and summing one turn:

± + β + γ + ... + ω = 2π

The interior angles formed by the sides AB and ZA, BC and BA, CD and CB,... ZA and ZY
are supplementary of ±, β, γ, ... ω. Therefore the sum of the angles of a polygon is:

( π ’ ± ) + ( π ’ β ) + ( π ’ γ ) + ... + ( π ’ ω ) = n π ’ 2π = ( n ’ 2 )π

n being the number of sides of the polygon. The deduction for the clockwise orientation of
the polygon is analogous with the only difference that the result is negative.


Definition of trigonometric functions and fundamental identities

Let us consider a circle with radius r (figure 6.2). The extreme of the radius is a
point on the circumference with coordinates (x, y). The arc between the positive X
semiaxis and this point (x, y) has an oriented length s, positive if counterclockwise and
otherwise negative. Also, the X-axis,
the arc of circumference and the
radius delimit a sector with an
Figure 6.2
oriented area A. An oriented angle
± is defined as the quotient of the arc
length divided by the radius2:
s
±=
r

Since the area of the sector is
proportional to the arc length and the
area of the circle is 2πr, it follows
that:

± r2 2A
±=
A= ”
r2
2

The trigonometric functions3, sine, cosine and tangent of the angle ±, are respectively
defined as the ratios:
y x y
sin ± = cos ± = tg ± =
r r x

and the cosecant, secant and cotangent as their inverse fractions:



2
With this definition it is said that the angle is given in radians, although an angle is a quotient of
lengths and therefore a number without dimensions.
3
Also called circular functions due to obvious reasons, to be distinguished from the hyperbolic
functions.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 55

r 1 r 1 x 1
csc ± = sec ± = cot ± =
≡ ≡ ≡
x cos ±
y sin ± y tg ±


being r related with x and y by the Pythagorean theorem:

r2 = x2 + y2

Then the radius vector v is:

v = x e1 + y e2 = r ( cos± e1 + sin± e2 )

From these definitions the fundamental identities follow:

sin ± 1
tg ± ≡ sin 2± + cos 2 ± ≡ 1 1 + tg 2 ± ≡ ≡ sec 2 ±
cos ± cos ±
2



If we take the opposite angle “± instead of ±, the sign of y is changed while x and r are
preserved, so we obtain the parity relations:

sin(’ ± ) ≡ ’sin ± cos(’ ± ) ≡ cos ± tg(’ ± ) ≡ ’ tg ±

The sine and tangent are odd functions while the cosine is an even function.
Look at the figure 6.2: β = π/2’± is the complementary angle of ± . If ± is higher
than π/2, the angle β becomes negative. On the other hand, if the angle ± is negative then
β is higher than π/2. The trigonometric ratios for β give the complementary angle
identities:

x y x 1
sin β = ≡ cos ± cos β = ≡ sin ± tg β = ≡
y tg ±
r
r


Angle inscribed in a circle and double angle identities

Let us draw any diameter PQ and any
Figure 6.3
radius OA (figure 6.3) in a circle with centre O.
Since OP is also a radius, the triangle POA is
isosceles and the angles OPA and PAO, which
will be denoted as ±, are equal. Because the
addition of the three angles is equal to π, the
angle AOP is π ’ 2± . The angle QOA is
supplementary of AOP, and therefore is equal to
2±, the double of the angle APQ. Let us draw
from A a segment perpendicular to the diameter
and touching it at the point N. By the definition of
sine we have:
RAMON GONZALEZ CALVET
56

NA NA PA
sin 2± = =
OA PA OA

The first quotient is sin± for the triangle PNA. If M is the midpoint of the segment
PA, then ¦PA¦ = 2 ¦MA¦ and the second quotient is equal to 2 cos± for the triangle
MOA:
MA
sin 2± = 2 sin ± ≡ 2 sin ± cos ±
OA

Through an analogous way we obtain cos 2± :

( )
PN ’ PO
ON PA
cos 2± = ≡ 2 cos 2 ± ’ 1 ≡ cos 2 ± ’ sin 2±
=
OA PA OA

Also this result may be obtained from the second fundamental identity. In order to obtain
the tangent of the double angle we make use of the first fundamental identity:

2 sin ± cos ±
sin 2± 2 tg ±
tg 2± ≡ ≡ ≡
cos2± cos 2 ± ’ sin 2± 1 ’ tg 2 ±

Finally, let us draw any other segment PB (figure 6.3). The angle BPQ will be
denoted as β. By the same arguments as above the angle BOQ is 2β. While the angle BPA
is ± + β, the angle BOA is 2± +2β : an angle inscribed in a circumference is equal to the
half of the central angle (angle whose vertex is the centre of the circumference) which
intercepts the same arc of circle. Consequently, all the angles inscribed in the same circle
and intercepting the same arc are equal independently of the position of the vertex on the
circle.


Addition of vectors and sum of trigonometric functions

Let us consider two unitary vectors forming the angles ± and β with the e1
direction (figure 6.4):

u = e1 cos ± + e 2 sin± v = e1 cos β + e 2 sin β

u + v = e1 (cos ± + cos β ) + e 2 (sin ± + sin β )
Figure 6.4
The addition of both vectors, u + v , is the
diagonal of the rhombus which they form, whence
it follows that the long diagonal is the bisector of
the angle ± ’ β between both vectors. The short
diagonal cuts the long diagonal perpendicularly
forming four right triangles. Then the modulus of
u + v is equal to the double of the cosine of the
half of this angle:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 57


±’β
u + v = 2 cos
2

Moreover, the addition vector forms an angle (± + β) /2 with the e1 direction:

±’β « ±+β ±+β
u + v = 2 cos + e 2 sin
¬ e1 cos ·
2 2 2

By identifying this expression for u + v with that obtained above, we arrive at two
identities, one for each component:

±+β ±’β
cos ± + cos β ≡ 2 cos cos
2 2
±+β ±’β
sin ± + sinβ ≡ 2 sin cos
2 2

In a similar manner, but using a subtraction of unitary vectors, the other pair of
identities are obtained:

±+β ±’β
cos ± ’ cos β ≡ ’2 sin sin
2 2
±+β ±’β
sin ± ’ sin β ≡ 2 cos sin
2 2

The addition and subtraction of tangents are obtained through the common
denominator:

sin ± sinβ sin ± cos β ± cos ± sin β sin (± ± β )
tg ± ± tg β ≡ ± ≡ ≡
cos ± cosβ cos ± cos β cos± cosβ


Product of vectors and addition identities

Let us see at the figure 6.4, but now we calculate the product of both vectors:

v u = ( e1 cos β + e 2 sin β ) ( e1 cos ± + e 2 sin ± ) =

= cos ± cos β + sin ± sin β + e12 (sin ± cos β ’ cos ± sin β )

Since u and v are unitary vectors, their product is a complex number with unitary
modulus and argument equal to the oriented angle between them:

v u = cos(± ’ β ) + e12 sin(± ’ β )

The identification of both equations gives the trigonometric functions of the angles
RAMON GONZALEZ CALVET
58

difference:

sin(± ’ β ) ≡ sin± cos β ’ cos ± sin β

cos(± ’ β ) ≡ cos ± cos β + sin ± sin β

Taking into account that the sine and the cosine are odd and even functions respectively,
one obtains the identities for the angles addition:

sin(± + β ) ≡ sin± cos β + cos ± sin β

cos(± + β ) ≡ cos ± cos β ’ sin ± sin β


Rotations and De Moivre™s identity

If v' is the result of turning the vector v over an angle ± then:

v' = v (cos ± + e12 sin ± )

To repeat a rotation of angle ± by n times is the same thing as to turn over an angle n± :

v'' = v ( cos ± + e12 sin± ) = v ( cos n± + e12 sin n± )
n




cos n± + e12 sin n± ≡ ( cos ± + e12 sin± )
n


This is the De Moivre™s identity4, which allows to calculate the trigonometric
functions of multiplies of a certain angle through the binomial theorem. For example, for n
= 3 we have:

cos 3± + e12 sin 3± ≡ (cos ± + e12 sin ± )
3




≡ cos 3 ± ’ 3 cos ± sin 2± + e12 ( 3 cos 2 ± sin ± ’ sin 3± )

Splitting the real and imaginary parts one obtains:

cos 3± ≡ cos 3 ± ’ 3 cos ± sin 2± sin3± ≡ 3 cos 2± sin± ’ sin 3±

And dividing both identities one arrives at:

3 tg ± ’ tg 3 ±
tg 3± ≡
1 ’ 3 tg 2 ±

4
With the Euler™s identity, the De Moivre™s identity is:

exp( n± e12 ) ≡ [exp(± e12 )]
n
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 59


Inverse trigonometric functions

The arcsine, arccosine and arctangent are defined as the inverse functions of the
sine, cosine and tangent. They are multivalued functions and the principal values are taken
in the following intervals:

π π
y = arcsin x ” x = sin y and ’ ¤ y¤
2 2

x = cos y and 0 ¤ y ¤ π
y = arccos x ”

π π
y = arctg x ” x = tg y and ’ < y<
2 2

y = arccot x ” x = cot y and 0 < y < π


From the definitions the parity of the inverse functions follow immediately:

arcsin x ≡ ’arcsin (’ x ) arccos x ≡ arccos(’ x )

arctg x ≡ ’ arctg (’ x )

Through the fundamental identities for the circular functions one obtains the
following identities5 for the inverse functions:

[ ] x
arcsin x ≡ arccos 1 ’ x 2 ≡ arctg
1’ x2


[ ] ® 
1’ x2
arccos x ≡ arcsin 1 ’ x ≡ arctg
2

x
 
° »

® 1
x
arctg x ≡ arcsin ≡ arccos 
1+ x2 ° 1+ x2 »

where the brackets indicate that the identity only holds for positive values.
From the complementary angles identities we have:

π π
arcsin x ≡ ’ arccos x arctg x ≡ ’ arc cot x
2 2


5
The only inverse circular function predefined in the language Basic is the arctangent ATN(X), so
these identities allows us to program the arcsine and arccosine.
RAMON GONZALEZ CALVET
60

Exercises

6.1 Prove the law of sines, cosines and tangents: if a, b and c are the sides of a triangle
respectively opposite to the angles ±, β and γ, then:

a b c
= = (law of sines)
sin ± sin β sin γ

c2 = a2 + b2 ’ 2 ¦a¦ ¦b¦ cos γ (law of cosines)

±+β
tg
a+b 2
= (law of tangents)
±’β
a’b
tg
2

Hint: use the inner and outer products of sides.

6.2 Prove the following trigonometric identities:

±’β β ’γ γ ’±
cos(± ’ β ) + cos(β ’ γ ) + cos(γ ’ a ) ≡ 4 cos ’1
cos cos
2 2 2

±’β β ’γ γ ’±
sin(± ’ β ) + sin(β ’ γ ) + sin(γ ’ ± ) ≡ ’4 sin sin sin
2 2 2

6.3 Express the trigonometric functions of 4± as a polynomial of the trigonometric
functions of ± .

6.4 Let P be a point on a circle arc whose extremes are the points A and B. Prove that the
sum of the chords AP and PB is maximal when P is the midpoint of the arc AB.

6.5 Prove the Mollweide™s formulas for a triangle:

±’β ±’β
cos sin
a+b a’b
2 2
= =
γ γ
c c
sin cos
2 2

6.6 Deduce also the projection formulas:

a = b cos γ + c cos β b = c cos ± + a cos γ c = a cos β + b cos ±

6.7 Prove the half angle identities:

± 1 ’ cos ± 1 ’ cos ± sin±
± 1 ’ cos ± ± 1 + cos ±
≡± ≡± tg ≡ ± ≡ ≡
sin cos
1 + cos ± sin± 1 + cos ±
2 2 2 2 2
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 61


7. SIMILARITIES AND SINGLE RATIO

Two geometric figures are similar if they have the same shape. If the orientation of both
figures is the same, they are said to be directly similar. For example, the dials of clocks are
directly similar. On the other hand, two figures can have the same shape but different
orientation. Then they are said to be oppositely similar. For example our hands are oppositely
similar. However these intuitive concepts are insufficient and the similarity must be defined
with more precision.


Direct similarity (similitude)

If two vectors u and v on the plane
form the same angle as the angle between the
Figure 7.1
vectors w and t, and they have proportional
lengths (figure 7.1), then they are
geometrically proportional:

± (u, v ) = ± (w, t )

u w u v ’1 = w t ’1
’
=

v t 

that is, the geometric quotient of u and v is equal to the quotient of w and t, which is a complex
number. This definition of a geometric quotient is also valid for vectors in the space provided
that the four vectors lie on the same plane. In this case, the geometric quotient is a quaternion, as
Hamilton showed.
The geometric proportionality for vectors allows to define the similarity of triangles.
Two triangles ABC and A'B'C' are said to be directly similar and their vertices and sides denoted
with the same letters are homologous if:

AB BC ’1 = A'B' B'C' ’1

that is, if the geometric quotient of two sides of the first triangle is equal to the quotient of the
homologous sides of the second triangle. Arranging the vectors of this equation one obtains the
equality of the quotients of the homologous sides:

AB’1 A'B' = BC ’1 B'C'

One can prove easily that the third quotient of homologous sides also coincides with the
other quotients:

AB’1 A'B' = BC ’1 B'C' = CA’1 C'A' = r

The similarity ratio r is defined as the quotient of every pair of homologous sides, which
is a complex constant. The modulus of the similarity ratio is the size ratio and the argument is
the angle of rotation of the triangle A'B'C' with respect to the triangle ABC.
RAMON GONZALEZ CALVET
62


A' B'
exp[± ( AB, A' B' ) e12 ] Figure 7.2
r=
AB

The definition of similarity is
generalised to any pair of polygons in the
following way. Let the polygons ABC...Z and
A'B'C'...Z' be. They are said to be directly
similar with similarity ratio r and the sides
denoted with the same letters to be homologous
if:

r = AB ’1 A'B' = BC ’1 B'C' = CD ’1 C'D' = ... = YZ ’1 Y'Z' = ZA ’1 Z'A'

One of these equalities depends on the others and we do not need to know whether it is
fulfilled. Here also, the modulus of r is the size ratio of both polygons and the argument is the
angle of rotation. The fact that the homologous exterior and interior angles are equal for directly
similar polygons (figure 7.2) is trivial because:


B'A' B'C' ’1 = BA BC ’1 ’ angle A'B'C' = angle ABC

C'B' C'D' ’1 = CB CD ’1 ’ angle B'C'D' = angle BCD etc.

The direct similarity is an equivalence relation since it has the reflexive, symmetric and
transitive properties. This means that there are classes of equivalence with directly similar
figures.
A similitude with ¦r¦=1 is called a displacement, since both polygons have the same
size and orientation.


Opposite similarity

Two triangles ABC and A'B'C' are
oppositely similar and the sides denoted with
the same letters are homologous if:

AB BC ’1 = ( A'B' B'C' ’1 )* = B'C' ’1 A'B'

The former equality cannot be arranged in a
quotient of a pair of homologous sides as we
have made before. Because of this, the
similarity ratio cannot be defined for the Figure 7.3
opposite similarity but only the size ratio,
which is the quotient of the lengths of any two
homologous sides. An opposite similarity is always the composition of a reflection in any line
and a direct similarity.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 63


AB BC ’1 = v ’1 A'B' B'C' ’1 v BC ’1 v ’1 B'C' = AB ’1 v ’1 A'B' ”


BC ’1 ( v ’1 B'C' v ) = AB ’1 ( v ’1 A'B' ’1 v ) = r

where r is the ratio of a direct similarity whose argument is not defined but depends on the
direction vector v of the reflection axis. Notwithstanding, this expression allows to define the
opposite similarity of two polygons. So two polygons ABC...Z and A'B'C'...Z' are oppositely
similar and the sides denoted with the same letters are homologous if for any vector v the
following equalities are fulfilled:

AB ’1 ( v ’1 A'B' ’1
v ) = BC ’1 ( v ’1 B'C' v ) = .... = ZA ’1 ( v ’1 Z'A' v )

that is, if after a reflection one polygon is directly similar to the other. The opposite similarity is
not reflexive nor transitive: if a figure is oppositely similar to another, and this is oppositely
similar to a third figure, then the first and third figures are directly similar. Then there are not
classes of oppositely similar figures.
An opposite similarity with ¦r¦=1 is called a reversal, since both polygons have the
same size both opposite orientations.


The theorem of Menelaus

For every triangle ABC (figure 7.4), Figure 7.4
three points D, E and F lying respectively on
the sides BC, CA and AB or their prolongations
are aligned if and only if:

AF FB ’1 BD DC ’1 CE EA ’1 = ’1

Proof ’ Let us suppose that D, E and F are
aligned on a crossing straight line. Let us
denote by p, q and r the vectors with origin at
the vertices A, B and C and going
perpendicularly to the crossing line. Then every pair of right angle triangles having the
hypotenuses on a common side of the triangle ABC are similar so that we have:

BF ’1 AF = q ’1 p CD ’1 BD = r ’1 q AE ’1 CE = p ’1 r

Multiplying the three equalities one obtains:

AE ’1 CE CD ’1 BD BF ’1 AF = 1

Taking the complex conjugate expression and changing the sign of BF, CD and AE, the theorem
is proved in this direction:

AF FB ’1 BD DC ’1 CE EA ’1 = ’1

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