# Monge's viewing method to prove Desargues' theorem

For instance, consider the triangle in space (see Figure 2.4). Assume that a triangle $$A,B, C$$ is projected to two different mutually perpendicular projection planes. The vertices of the triangle are mapped to points $$A',B',C'$$ and $$A'',B'',C''$$ in the projection planes. Furthermore, assume that the plane that supports the triangle contains the line $$l$$ in which the two projection planes meet. Under this condition the images $$ab'$$ and $$ab''$$ of the line supporting the edge $$AB$$ will also intersect in the line $$l$$. The same holds for the images $$ac'$$ and $$ac''$$ and for $$bc'$$ and $$bc''$$. Now let us assume that we are trying to construct such a descriptive geometric drawing without reference to the spatial triangle. The fact that $$ab'$$ and $$ab''$$ meet in $$l$$ can be interpreted as the fact that the spatial line $$AB$$ meets $$l$$. Similarly, the fact that $$ac'$$ and $$ac''$$ meet in $$l$$ corresponds to the fact that the spatial line $$AC$$ meets $$l$$. However, this already implies that the plane that supports the triangle contains $$l$$. Hence, line $$BC$$ has to meet $$l$$ as well and therefore $$bc'$$ and $$bc''$$ also will meet in $$l$$. Thus the last coincidence in the theorem will occur automatically. In other words, in the drawing the last coincidence of lines occurs automatically Source: Page-39 and 40 of Perspectives on Projective Geometry A Guided Tour Through Real and Complex Geometry

I am totally confused on whatever is going on in the above paragraph.

Here is what I understand so far:

Monge's method involves projecting a 3-d Object onto some planes and then reconstructing the object back in 3d from the projections. An analogy I found helpful for thinking about this is how we can construct a vector by projecting it's component with respect to a set of standard bases.

### What I don't understand:

1. What is ab' and ab'', bc and bc''? It is not defined in the passage..
2. Where is the original triangle ABC in the picture? has it collapsed onto a one dimensional line in our viewing prespective?
• I haven't an answer but this is consistant with the applied field of "descriptive geometry" Monge has developed. Sep 14 at 17:07

My best guess is that the thick middle line is $$\ell$$, and that it is on a plane seen head on, and containing triangle $$ABC$$, as you conjectured.
There are also two planes going through $$\ell$$, one above and one below the first plane. Although they are described as mutually perpendicular, I don't think that's necessary. $$ABC$$ is parallel projected to these planes to form $$A'B'C'$$ and $$A''B''C''.$$
$$ab'$$ should be $$A'B'$$, $$ab''$$ should be $$A''B''$$ and so on.