First of all, in order to make sense of the question one has to specify the axiomatic system via which the (Euclidean) plane is defined. For the purpose of this answer, I assume that you are interested in Hilbertian axiomatization of the Euclidean geometry. (There are alternatives, such as Birkhoff's and Tarski's sets of axioms.) Hilbert's axiomatization has two equivalence relations built in: congruence of segments and congruence of angles. If you want to change Hilbert's notion of congruence to something else, you are leaving the realm of Hilbert's axioms and, hence, you would have to list your personal axioms and then ask a new question.
Given this, the answer is that, by the definition, two segments have the same length iff they are congruent, the congruence class of a segment $AB$ is denoted $[AB]$.
This, of course, is not entirely satisfactory since we are used to the idea that length should be a real number and not an element of some unfamiliar quotient set (coming from some equivalence relation). I suspect that this is where your question is coming from.
It turns out that one can (assuming all but the continuity axioms of Hilbert) define an ordered field $F$ whose elements are the congruence classes, you can find a detailed construction in the book:
R. Hartshorne, "Geometry, Euclid and Beyond", Springer Verlag, 2000.
One needs the two continuity axioms (of Hilbert's axiomatic system) in order to prove that $F$ is isomorphic to the field of real numbers.
Note. In contrast, if you accept Birkhoff's axiomatization, then the Euclidean plane $E$ comes equipped with a distance function $d: E\times E\to {\mathbb R}_+$. The length of a segment $AB$ is simply the real number $d(A,B)$. A congruence of this geometry is an isometry, i.e. a map $g: E \to E$ preserving the distance function. Two segments $AB, CD$ are congruent in this sense iff there exists an isometry $g$ such that $g(AB)=CD$. Thus, two congruent segments have the same length, again, just by the definition. A good reference for this is:
E. Moise, "Elementary Geometry from an Advanced Standpoint", Pearson, 1990.
Moise sketches the relation between the two (Hilbert and Birkhoff) sets of axioms and explains, for instance, how to add congruence classes in Hilbert's plane. But he does not verify any of the ordered field axioms (for instance, he does not explain how to multiply the congruence classes).
Edit. I think, I understand now the source of your confusion, it is hidden in the article an. The point is that Hilbert's axiomatic system does not just contain an equivalence relation, called congruence, but a very specific one, which you can find for instance in either one of the references I gave above. If you change Hilbert's segment congruence equivalence relation to a different equivalence relation, you obtain a different axiomatic system. It appears that you are interested in replacing the standard segment congruence equivalence relation with the similarity equivalence relation. This equivalence relation has exactly two equivalence classes, namely, one represented by degenerate segments $AA$ and one represented by nondegenerate segments $AB$, $A\ne B$. This is all fine, but now you are asking about "length" in relation to this equivalence relation. For this, you have to specify what do you mean by the "length" since the latter is not a part of Hilbert's axiomatic. There are two meaningful choices:
a. Use the notion of length $[AB]$ as the equivalence class of the segment $AB$ with respect to the standard notion of congruence. This notion is clearly not preserved by the similarity: Similar segments, in general, do not have the same length. This is quite immediate (you can use the fact that there are infinitely many congruence classes of segments in the traditional sense and only two similarity equivalence classes). Thus, the answer to your question is negative in this case.
b. Use the notion of "length" of $AB$ as the similarity equivalence class of the segment $AB$. Then your notion of "length" takes exactly two values and is, but the definition, preserved by the similarity equivalence relation. Very few (if any) people, however, will accept this notion of "length" as useful, for instance, because it does not take values in the ordered field (the field of two elements cannot be ordered).
Best Answer
This is equivalent to say all the basic operations on $A$ can be defined as over the equivalence classes of $\theta$ by $f([x_1], \cdots, [x_n]) := [f(x_1, \cdots, x_n)]$ -- using the substitution property, it can be shown that this is well-defined. Therefore, we may define $A/\theta$ as a quotient algebra, with the natural map $A\rightarrow A/\theta$ being a homomorphism.
On the other hand, if we have a homomorphism of algebras $\rho: A\rightarrow B$, and define $x\theta y \Leftrightarrow \rho(x)=\rho(y)$, then $\theta$ is a congruence relation on $A$.
This shows that congruence relation is equivalent to quotient algebra, and surjective morphism. This is where universal algebra is different from group/ring theory: In groups, a congruence relation is equivalent to a normal subgroup, and in rings, a congruence relation is equivalent to an ideal. However, in universal algebra, kernels is not necessarily defined, and different equivalence classes of a congruence relation may not even have the same cardinality, so they are in general not necessarily cosets of something.