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## Formal Proof of Invariant Incidence

There is a risk of circularity here, as the Theorem of Desargues depends on perspectivities,
and in what follows, we have perspectivities depending on the Theorem of Desargues.
It is to resolve.
The two “intervertical” lines, p and m, incident in G, are common to both pairs of Desargues triangles, blue and red, so the pairs share a “perspector”, and  a “perspectrix”. A similar exercise undertaken with Δacb1 (ΔPMU) and Δacb1 (ΔPMU) would lead to the same point, G, so one may say that the incidences on which a ruler depends are independent of particular constructions, provided they all begin from the same four of the base points.

Formality aside, the ‘barred’ and ‘unbarred’ constructions are ‘shadows’ or perspectivities, of each other, from centre G, so
are equivalent or identical.
We have blue, linewise Δxab1 (which is, pointwise, ΔPMS) and blue, linewise Δxab1 (which is, pointwise, ΔPMS). Because
• side x is incident with side x in point X
• and side a is incident with side a in point A
• and side b1 is incident with side b1 in point B
• and X, A and B are collinear,
the blue triangles are a Desargues pair.

Lines p, m and s, joining corresponding vertices, therefore converge in point G.
We have red, linewise Δycb1 (which is, pointwise, ΔPMN) and red, linewise Δycb1 (which is, pointwise, ΔPMN). Because
• side y is incident with side y in point Y,
• and side c is incident with side c in point C,
• and side b1 is incident with side b1 in point B,
• and Y, C and B are collinear,
the red triangles are a Desargues pair.

Lines p, m and n, joining corresponding vertices, therefore converge in point G.
←Tutorial page                  ← Related page                  ← Comparison 2 - Desargues Mode               ← Comparison 7 - Replication “in place”.               Five Planes ↑