How can affine plane extended of projective plane?

In the book of Euclidean and Non-Euclidean Geometry by Greenberg, it is given that

DEFINITION: Lines l and m are parallel if they are distinct lines and no point is incident with both of them.

DEFINITION: A projective plane is a model of incidence geometry hav­ing the elliptic parallel property (any two lines meet) and such that every line has at least three distinct points lying on it (strengthened In­ cidence Axiom 2).

DEFINITION: An affine plane is a model of incidence geometry hav­ing the following Euclidean parallel property:

For all line $l$, and for all points $P$ that are not incident to $l$, there exists a unique line $m$ s.t $P$ is incident to $m$ and $m$ and $l$ are parallel.

[...] So the idea in extending an affine plane to a projective plane is to add enough new "points at infinity" so that all lines parallel to any given line will now meet at one such point.

However, by definition two parallel line cannot have a common point which is incident to both, and when we add "new points at infinity" to make all line meet at a single point, we will not have any parallel lines anymore, so we cannot be working in affine geometry anymore.

Moreover, isn't the definition of affine plane same as the parallel postulate of Euclid ?

Edit: In case I misunderstood someting, here is the exact formulation: ($\exists !$ means exits uniquely)

• ...Yes, when you extend your affine plane, you now have a projective plane instead of an affine plane. – Eric Wofsey May 18 '18 at 6:36
• @EricWofsey Then the word "extend" misused in here because I always understand that word as we are adding new things without spoiling the existing things, whereas in this case, we are transforming the existing geometry to another. – onurcanbektas May 18 '18 at 6:39
• Would you say that $\mathbb{C}$ is not an extension of $\mathbb{R}$, because it "spoils" the fact that $-1$ has no square root? – Eric Wofsey May 18 '18 at 6:41
• In $\mathbb{R}$, there is an element that has no square root. In $\mathbb{C}$, there isn't. How is that any different? It's not as though the parallel lines stopped existing; they just stopped being parallel, the same way $-1$ stopped having no square root. – Eric Wofsey May 18 '18 at 6:44
• We can read it this way : we add a new axiom : the elliptic parallel property, i.e. any two lines meet. This axiom "cut-off" the Euclidean plane, i.e. it is no more a model of the new theory, because it does not satisfy the new axiom. Then we consider a new "environment" : the Euclidean plane with the addition of the new points at infinity : now the above axiom is satisfied. – Mauro ALLEGRANZA May 18 '18 at 6:58

You can however use the projective plane as a model for the affine plane, as long as you keep track of which line is the line at infinity. So for example, the operation “draw parallel to $l$ through $P$” from affine geometry would become “intersect $l$ with the line at infinity, then connect that point of intersection with $P$”. Different vocabulary, but in 1:1 correspondence so both describe the same mathematical structure.