Historically First Geometric Presentation of Random Acute Angle’s Trisection-Euclidean Solution

(With the use only of compass & non measuring ruler)

By Giorgios (Gio) Vassiliou (Greece) – Visual Artist – Founder and Inventor of Transcedental Surrealism in visual arts – Architect

“Δος μοι πα στω και ταν γαν κινάσω!”
“Give me a place to stand and i will move the Earth!”

Αrchimides of Syracuse



First we have drawn a random angle. Then we tranvert this angle to an inscribed one of a circle. We draw the circle (O, OA’). We see the points A’ and B’ are the intersections of angle’s sides with the circle.We draw the bisector OD of angle A’OB’.


From the center point M of OD we draw circle (M, MD), and the intetsections of the angle with the new circle are points A, B. Now we will work with angle AOB. Continiously we extend from point O the angle’s sides AO, OB and we create a new central angle. Now we draw the circle (O, OA”) r=OA”=MD. The new central angle A”OB” (as an extension) is equal to angle AOB. And angle A”OB” is central angle of inscribed angle AOB.


angle A”OB”(central) =angle AOB(inscribed)


arch AOB(inscribed angle) =2*arch A”OB”(central angle)

Also D’ is the bisector’s OD extensional point on angle A”OB”. OD’ is also the bisector of angle A”OB”.


We draw the line AA”. In this point you will allow me, to call AA’ as the “DIVINE LINE” , because of its crusial importance. Then from center O we draw a parallel line ZZ’//AA”. This line will be called the “HUMAN PARALLEL”.

As we see, now we have to rotate the angles A”OB” and AOB, at an angle φ=angle D’OZ’. So point Z’ now will meet point D’ and Z point will meet point D.

So the whole geometric structure will rotate at angle φ.


This very geometric structure that leads to trisection, from now on it will be known by the name of “TRISECTOR”!

Now the solution has reached the end and we have the following equalties.

a) AA”//ZZ’

b) arch A”Z’=1/2* arch ZB ( because angles A”OD’ & DOB are vertical between them and also angle A”OB” is the central angle and ZOB the inscribed angle of it. So arch ZB =2*archA”D’.

The same is applied for arches Z’B” and AZ. So arch AZ=2*archZ’B”)

c) arch A”Z’=arch AZ (they both are arches between parallel chords of equal circles)


arch AB= arch AZ+ arch ZB=>

arch AB= arch A”Z’+ arch ZB=>

arch AB= arch A”Z’ +2*arch AZ’=>

arch AB=3*arch A”Z’

So we have….

arch A”Z’=1/3 arch AB

& arch AZ =arch A”Z’=1/3 arch AB


Date: Wednesday 7 April 2021

Dedicated to the memory of my beloved parents Spiros & Stavroyla

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