Comparison 9, of Skew Rulers →

## Formal Proof of Invariant IncidenceThere 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 Δacb_{1 } (ΔPMU) and
Δ_{1
} (Δ ) 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‘mutually shadowed’ intervals
are equivalent or identical. |

We have blue, linewise Δxab_{1}
(which is, pointwise, ΔPMS) and blue, linewise Δ_{1} (which
is, pointwise, Δ ).
Because
- side
**x**is incident with side in point**X** - and side
**a**is incident with side in point**A** - and side
**b**_{1}is incident with side_{1}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 Δycb_{1} (which is, pointwise, ΔPMN) and red,
linewise Δ_{1} (which
is, pointwise, Δ ).Because
- side
**y**is incident with side in point**Y**, - and side
**c**is incident with side in point**C**, - and side
**b**_{1}is incident with side_{1}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. |