I'm following along the book "Basic Geometry" by Birkhoff and Beatley. It contains a proof for the Pythagorean Theorem:
In any right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides; and conversely.
Given triangle ABC in which $\angle C = 90°$. To prove: $a^2 + b^2 = c^2$
...
Converse: Given lengths $a, b, c$ such that $a^2 + b^2 = c^2$. To prove: there is one and only one right triangle having these lengths for sides.
Proof: Construct a triangle ABC such that $AC = b$, $\angle C = 90°$ and $CB = a$.
Then $a^2 + b^2 = c^2$. (Why?)
But $c^2 = a^2 + b^2$ (Given).
Therefore $(AB)^2 = c^2$ and $AB = c$
So there exists at least one right triangle with sides $a, b, c$. If more than one such triangle exists, these triangles must be equal, for it they are unequal, they violate Principle 8.
Principle 8 is SSS congruence.
I understand that if there is more than one triangle that fullfills $a^2 + b^2 = c^2$ for the same $a, b, c$ then these triangles must be equal.
What I don't understand is why the converse proof from above shows that there is at least one such triangle.
The claimed converse is the following statement: If the three sides of a triangle $\triangle$ satisfy $a^2+b^2=c^2$ then the triangle has a right angle at $C$.
For the proof we draw an auxiliary right triangle $\triangle'=A'B'C'$ with $|A'C'|=b$, $|B'C'|=a$, and a right angle at $C'$. There is no question that such a triangle exists. Using the already proven part of Pythagoras' theorem we then can say that $|A'B'|^2=a^2+b^2=c^2$, hence $|A'B'|=c$. This shows that $\triangle'$ has the same side lengths as $\triangle$. By "Principle 8" it is then congruent to $\triangle$. This implies that the angles are equal as well; in particular the angle at $C$ is $90^\circ$.