MATHEMATICS AND ORIGAMI

Mathematics and Origami

9.9 TRISECTION OF ANY ANGLE ABC

∆BKL = ∆KLF (symmetry) ; BL = LF (symmetry)

∆BLD = ∆KLF (one side and angle, congruent)

so ∆BKL = ∆BLD, that is: ang JBF = ang FBC

Moreover: as GI = IB , HJ = JF (symmetry)

Said symmetry with respect to ED produces the right angles marked in I y J.

Then ∆HBJ = ∆FBJ, that is: ang ABJ = ang JBF

therefore: ang ABJ = ang JBF = ang FBC =

3

1

ang ABC

It is impossible to achieve that simultaneous coincidence by means of rule, square / set

square and compass, nor even by CAD. The computer can help in a try and cut process, but

that ́s all. On the contrary, folding is the unique way to integrate trying, intuition and pre-

cision.

A

B D C

I

G

E

H

J

K F

L

A

E

G

I

B D C

F

J

H

A A

G

B

B C C

=

=

I

We have trisected angle ABC, associated to its corresponding square. Of course, any

angle can be related to a square.

The reader may have observed that we came across the solution by simultaneously

folding under two different conditions. Something similar was shown when resolving the third

degree equation (Points 7.11; 7.14; 7.14.5). Now it is pertinent to say the following:

Process will be as follows:

1- To get two straight lines parallel to a side (e.g. BC) under

the condition to be equidistant as the figure shows. Thus we get

points G and I.

2- Produce valley fold ED such that B will lie on F (over

lower parallel) and, simultaneously, G on H (over ray BA). As

a consequence of the symmetry (axle ED), I will lie on the new

point J.

3- Rays BJ and BF trisect Ang ABC. Justification: