 Perpendicular bisector and midpoint

As in the Euclidean case the hyperbolic perpendicular bisector is the geometric place of the points of the plane that are equidistant of two given points. To construct the hyperbolic perpendicular bisector we will use that the inversions with center in a point of the boundary line are isometries for this model.

The midpoint of a segment is the point that belongs to the segment and equidists of each endpoint. Therefore, the midpoint is a point of the perpendicular bisector, in
fact, the intersection of the perpendicular bisector with the segment. That is why we make an only construction for both objects.

We will follow the following steps:
1. Construct the hyperbolic segment from the two given points, C and D.
2. Plot the Euclidean line that goes through both points.
3. Consider the intersection of this line with the boundary line. We designate this point for P.
4. Plot the circumference of center P and radius √(d(P,C).d(P,D)). This circumference will be considered as the inversion circumference.
5. Consider the intersection between this circumference and the boundary line.
6. Consider the intersection between this circumference and the hyperbolic segment constructed in the first step.
7. Plot the arc of circumference that passes through an intersection of the fifth step, the intersection of the sixth step and the other intersection of the fifth step. We affirm that this arc of circumference is the hyperbolic perpendicular bisector and the intersection point of the sixth step is the hyperbolic midpoint of the given hyperbolic segment. To prove that this construction really gives us the perpendicular bisector and the hyperbolic
midpoint, it is necessary to prove that the distance between one of the given points and the point that we suppose midpoint it is the same that the distance between the other given point and the midpoint.
As the isometries preserve the distances we have that these two distances are equal if there is an inversion with center to the boundary line that transforms one of the given points in the other one and leaves fixed the created point. But this inversion is precisely the circumference that we have constructed in fourth step since it has center in the boundary line, goes through the midpoint (we have constructed the midpoint from the intersection of this circumference) and exchange the given points. The last affirmation is true because we have that both points are lined up with the center of the circumference, for construction, and the square of the radius of the inversion circumference is precisely the product of the distance from the center, P, in each point. So, the intersection point of the sixth step is the midpoint.
Moreover, we have that all the points of the arc plotted in seventh step are points of the perpendicular bisector since all the points of the arc are at the same distance from the two given points. This is because the same inversion brings the first point to the second and fixes all the points of the perpendicular bisector. List of tools
Hyperbolic geometry