Hyperbolic equidistant

With this tool we will construct the equidistant from hyperbolic lines that are Euclidean semicircumferences. The hyperbolic lines that are also Euclidean lines will not be considered because the hyperbolic line tool does not work in this situation. Anyway the reader can create easily the equidistant tool for the case where the hyperbolic line is an euclidean line since in the steps we will follow to construct the equidistant we will construct the hyperbolic equidistant for a perpendicular line to the boundary line.

In Euclidean Geometry the equidistant from a line is another line but in Hyperbolic Geometry we do not obtain a hyperbolic line. To construct the equidistant we have based in the theory that we used to ensure its existence (Workshop of Hyperbolic Geometry.pdf, in Catalan).

To construct the equidistant we need two points, that will determine the hyperbolic line, and a distance, which we want the equidistant to be of the hyperbolic line. As we will use distances we will need to fix the axes, but we know that the result will not depend on where we have fixed them. So, the steps that we follow are the following ones:

- Plot the hyperbolic
line that passes through the two given points.

- Consider the following points: one of the two intersection points between the hyperbolic line just constructed and the boundary line and intersection point between one of the perpendicular line to the boundary line that passes through one of the given points. Plot the Euclidean circumference with center in the first described point and passes through the second point. This will be the inversion circumference from which we will transform the hyperbolic line plotted in the first step to a perpendicular line.
- Consider the intersection point between the line plotted in the first step and the Euclidean circumference plotted in the second step. Plot the perpendicular line to the boundary line that goes through the considered point. This is the line that we would obtain from the inversion of the hyperbolic line plotted in the first step.
- Plot the hyperbolic circumference with center at any point of the perpendicular line and radius the given distance. We have taken the intersection point between the original hyperbolic line and the inversion circumference. Many other points could be chosen but with this we can assure that the hyperbolic circumference will cut the original hyperbolic line.
- Consider the intersection point
between the perpendicular line plotted in the third step and the
boundary line and consider the intersection points between the
hyperbolic line plotted in the first step and the Euclidean
circumference plotted in the fourth step. Plot the Euclidean rays with
origin in the first considered point and passing through the second
considered intersection. These two rays are the two equidistants to the
hyperbolic line plotted in the third step.

- Consider the two intersections of the two rays plotted in the former step with the inversion circumference, plotted in the second step. These points are fixed by the inversion.
- Plot the Euclidean circumference obtained from the inversion of one of the two lines plotted in fifth step. For the properties of the inversions we can plot the circumference that passes through the two intersection points obtained in the former step, by the line chosen, and by the center of the inversion. To plot this circumference we will not use the tool that gives us the arc from three points since we need the whole circumference to consider only the arc that belongs to the half-plane. So, it is necessary searching the center. To make it, we have to consider two of the three Euclidean segments that we obtain joining the three points. For these two segments we plot their Euclidean perpendicular bisector. The intersection point will be the center. We will always be able to make this construction since we are now thinking about Euclidean Geometry.
- Consider the intersection between the circumference constructed in the former step and the boundary line.
- Plot the arc of circumference that has origin in one of the two former points, passes through the intersection point of the sixth step that belongs to the Half-Plane and finishes in the other intersection point.
- Repeat the construction as in the three former steps but for the other line. In this way, we have obtained two equidistant at the fixed distance to the given hyperbolic line.

Hyperbolic geometry