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## The KEY Once the four circles have been identified by their respective diameters in decreasing gold­en pro­por­tion, let's now move on to the first initiative, which con­sists in tracing from the vertical point `V` on the larg­er cir­cum­fer­ence, of diameter = `1`, a line tan­gent to the circumference Φ, which will meet the primary at point `A`. With or without al­ge­bra­ic proof – I pro­duced it here in 2009 – we will im­me­di­ate­ly find out some interesting aspects:
1. drawing the symmetrical `VB` line to `VA` and
joining `A` with `B` we will have drawn a triangle whose base `AB` is tangent to the circle Φ3;
therefore we would have obtained `B` also fol­low­ing the reverse process.
2. a mirror line at `VA` halves the side at the point `X` which is tangency of `VA` to the circle Φ; and reaches exactly the point of tangency at Φ3 in the middle of `AB`; in particular a line parallel to `AB` from the point `X` is tangent to Φ2, perhaps I should add 'obviously'; but the best is yet to come
3. The sum of `AV` and `VB` is equivalent to `AB×`Φ, i.e. the base of this es­pe­cial tri­an­gle is the Golden Section of the sum of the sides; the supreme bal­anc­ing factor in the structure of the ordered physical universe; this had since long led me to define it as the Great Golden Triangle par excellence. Driven by these premises, I forwarded to an in-depth study that proved to be de­ci­sive; and now the time has come to grap­ple with the num­bers.

Given the diameter of the pri­ma­ry circle as a unit of meas­ure­ment, therefore with a val­ue of `1.0`, the cir­cum­fer­ence will be `= π` and the quad­rant arc `VQ = π/4` (in a linear sense, as opposed to angular meas­ure­ment in radians).

In this case the `VD` height of the tri­an­gle will be `VC+CD` ie
`0,5 + `Φ`³/2` [0,2360679774997896964/2… = 0,1180339887498948482…+ 0,5], which in our construct turns out to be the golden section Φ of the diameter = 1:
`0,618033988749894848204586834365638117720309179805762862135…`

To measure `VA` it will be neessary according to the theorem of Pythagoras:
`AD = √AC² – CD²`