Desargues Theorem Proof Using Homogeneous Coordinate

[wawar05]

Before I ask the question, let me remind that desargues theorem states :

if two triangles are perspective from one point then they are perspective from one line

I'd like to ask whether the order of the steps of the proof I did is correct or not. Since I saw the proof from an article but it only provided the image of the triangles. The proof relates to Projective Geometry

I did the proof of the theorem using homogeneous coordinate which starts with four coordinates D (1,1,1) as the perspective point, A (1,0,0), B (0,1,0), and C (0,0,1). Then ABC forms triangle. Then I extend line DA, DB, and DC and take respectively arbitrary point A' (1,a,a), B' (b,1,b), and C' (c,c,1). Then A'B'C' forms triangle

nb: I just show the beginning steps, since the remains is only algebraic problem and I am not confused on it.

Probably you know why in the article the second triangle has such coordinates A' (1,a,a), B' (b,1,b), and C' (c,c,1)?

additional questions: why (1,1,1), (1,0,0), and (1,a,a) can be in one line?
why (1,1,1), (0,1,0), (b,1,b), can be in one line?
and why (1,1,1), (0,0,1), and (c,c,1) can be in one line?

thank you very much.

 

[Eynstone]

Your approach is close to that of the standard proof using projective geometry. You can check for collinearity by determinants.