We have the following axioms:
Clairaut's Axiom. If $AB$, $AC$ and $BD$ are segments such that $AC$ and $BD$ are congruent and both perpendicular to $AB$, then the angles on $C$ and $D$ are both right. In other words, $\square ABDC$ is a rectangle.
Clavio's Axiom. The set of points that are equidistant to a given straight line is a straight line.
I am having issues to prove that Clavio implies Clairaut, in fact I have an idea to prove it on Hilbert plane, but it does not use Clavio.
It consists on that, if we have three segments two of them congruent and perpendicular to the other, then the line which connect the middle points of two congruent segments is perpendicular to both. Knowing this, we can extend the segments $AC$ and $BD$ of Clairaut's hypothesis to another points $C'$ and $D'$ such that $C$ and $D$ are the middle points of $AC'$ and $BD'$, so we have that $CD$ is perpendicular to $AC$ and $BD$ just as affirm Clairaut's axiom.
The problem here is that I am not using Clavio's axiom and I don't know if there is another way. Possible ideas would be appreciated.