
> 
> Is it possible to construct the midpoint of a segment in the hyperbolic plane 
>  using the *set square* only?
> 
> 
> 


With the *set square* one can 


* draw the line through the given two points and
* drop the perpendicular from the given point to the given line.




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The following construction produce the point $X'$ which is centrally symmetric to the point $X$ with respect to point $O$. 


1. Draw line $(OX)$ and let $m$ be the line perpendicular to $(OX)$ through $O$.
2. Draw yet two perpendicular lines $l$ and $l'$ through $O$.
3. Find the foot point $Y$ of $X$ on $l$.
4. Draw the line through $Y$ perpendicular to $m$ and let $Z$ be its intersection with $l'$.
5. Finally, $X'$ is the footpoint of $Z$ on $l$.



Mid point with set square?
Finally, I realized that the answer is NO.


Let us think of hyperbolic plane as about subset of projective plane for $\mathbb R^{2,1}$.


Note that in this model, the set square tool is the same as hyperbolic-cross-product tool;
i.e. for any two vectors $u$ and $v$ 
you can construct the vector 
$$w=J(u\times v),$$
where
$$ J\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x\\y\\-z\end{pmatrix}$$


If the end points of the segment are given by rational vectors then so are all the hyperbolic-cross-products. 
On the other hand, it is easy to construct two rational vectors such that their hyperbolic bisector goes in an irrational directin. 
Hence the result follows. 




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In **spherical geometry**, the same question has also answer NO.
It is easier to visualize since in spherical geometry the *set square tool* is equivalent to the *cross-product tool*.


