Jesús de la Peña Hernández

and rotate it 180º around O, Fig. 2 of Point 7.13 is obtained. Both figures give the same results

for t:

a = 51.2721º b=-23.9909º g = -60.9719

t 1 = tg a = 1.2469 t 2 = tg b = -0.445 0 t 3 = tg g = -1.8019

Any of them satisfy the equation

t^3 +t^2 − 2 t− 1 = 0

Fig. 4 justifies the association of Fig. 3 to the 3rd degree equation:

∆IBD; ∆EGF are similar:

GE

FG

ID

BD

= ;

1

2

2 −

=

ED

BD

(1)

∆IBC; ∆CEF are also similar:

BD

IC

GF

EC

= ;

BD

ED CD +CD

=

− 2

2

(2)

Equalising

2

BD

in (1) and (2):

ED CD

CD

ED −

+

=

−

2

1

2

;

1

2

+

=

ED

CD (3)

In ∆IBC we also have (t = tg a):

BD=IDtgα= 2 t ; Ang.BCI = 180 – 2a ; BD=DCtg() 180 − 2 α

Then:

tg 1

2 tg

2 tg (^22)
−
=− =
α
α
α
DC
t DC ; t^2 − 1 =DC (4)
In ∆EGF, Ang.FEG = a, so:
GE
GF
t= ;
1
2
−

ED
t (5)
Expressions (3), (4), (5) form a t parametric system that will allow us to obtain the 3rd degree
equation we are after:
F
I
4
B
D CG E