Do Tarski's (geometry) axioms imply that all zero segments are congruent? Tarski's axioms are an alternate formalization of geometry (similar to axiom sets of Euclid and later Hilbert). Do these axioms imply:
$$\forall\; x,y\in \text{points},\; x x\equiv y y?$$
If yes, what  is the proof? My feeling is that the proof must use Tarski's Identity of Congruence
    $$xy\equiv zz \rightarrow x = y,$$
but I am unable to find a proof.

This seems surprisingly tricky.  Identity of Congruence isn't enough by itself since it goes in the wrong direction.
I want to use the Five Segment Axiom.  Let me call your two points $p,q$ instead, to avoid a conflict with Wikipedia's notation.  If $p=q$ then we are done by reflexivity, so assume $p \ne q$.  Set $u=z=x'=p$ and $u'=z'=x=q$.  Let $y=y'$ be the midpoint of $pq$ (it takes some more work to prove that it exists), so $Bpyq$ and $py \equiv qy$.  Now we verify that the hypothesis of the Five Segment Axiom is satisfied, and conclude $zu \equiv z'u'$ which is to say $pp \equiv qq$.
