How is circumscribing a triangle equivalent to the Parallel Postulate?

Question

This question states that one of the statements equivalent to the parallel postulate (Euclid 5) is "Every triangle can be circumscribed". The Wikipedia page on Tarski's Axioms lists three variants of the Axiom of Euclid, one of which is "Given any triangle, there exists a circle that includes all of its vertices." Another discourse on Tarski's Axioms I have read has that for any three non-collinear points there is a point equidistant from all three. All three of these versions are stating the same thing, just in different words.

If this statement is equivalent to Euclid 5, then Euclid's proof of triangle circumscription should depend on Euclid 5. If it can be done without Euclid 5, then either (a) it isn't equivalent to Euclid 5, or (b) Euclid 5 can be proven by the other axioms.

However, when I look up Euclid Book IV, Prop 5, he describes how to circumscribe a triangle by constructing the bisectors of two sides to find the circumcenter at their intersection. His proof does not invoke Euclid 5 directly or indirectly that I can see.

This appears to contradict my assertion above about if circumscribing a triangle is equivalent to Euclid 5, then Euclid's proof that triangles can be circumscribed has to invoke Euclid 5.

Obviously, I'm missing something here, but I don't see what it is. How can the claim that "every triangle can be circumscribed" is equivalent to Euclid 5, and Euclid's Proposition IV.5 be rectified?

Answer 1

A good way to demonstrate such things is to try them in a model of non-Euclidean geometry. My favorite is the Poincaré Disk. In this model, all "points" live inside a particular circle (dashed in the images below), and "lines" are either diameters of that circle or arcs of circles orthogonal to it.

In the figures, $\triangle OAB$ has vertex $O$ at the center of the universal circle. Sides $\overline{OA}$ and $\overline{OB}$ lie on diameter-lines. $M$ and $N$ are midpoints of those sides, and perpendiculars through them are orthogonal-circle-lines. (Inversion the $M$-circle exchanges $O$ and $A$; likewise, inversion in the $N$-circle exchanges $O$ and $B$. This is analogous to the Euclidean property reflection in a perpendicular bisector line exchanges two vertices of a triangle.)

As shown below, the perpendicular bisectors at $M$ and $N$ need not meet. In the first figure, they do; in the second figure, they meet on the border of the universal circle (ie, "at infinity"), which is a special case; in the third figure, they don't meet at all.

Since the perpendicular bisectors need not meet, a circumcenter need not exist. Consequently, guaranteed circumscribability requires Euclidean-ness, so the Parallel Postulate lurks somewhere behind the property.

A further note ... It happens that circles in hyperbolic geometry are represented by Euclidean circles in the Poincaré model (although the centers don't match). In the figure, the yellow circle is the (Euclidean) circle through $O$, $A$, $B$. It corresponds to a valid hyperbolic circle only when it lives inside the universal circle, as in the first figure. With the second figure, the yellow circle is "tangent at infinity", giving a hyperbolic "horocycle", which doesn't count as a hyperbolic circle. In the third figure, the yellow model circle extends beyond the universal circle, so it cannot correspond to a valid hyperbolic circle.