Hyperbolic bisector

As in the Euclidean Geometry we can trace the angle bisector. To use this tool it will be necessary to give three different points. The order will be important since with these three points we will construct an angle with the vertex in the second given point.
The hyperbolic bisector has to divide an angle into two equal parts. To construct it we will use the Euclidean bisector. We will also use the following fact. Given two different Euclidian rays lying on the same line and with the same origin, if we want to decide wich ray is nearer from a fixed point we can trace the perpendicular line for that point. The ray the perpendicular line cuts will be the nearest. It is because the Euclidean distance from an exterior point to a line is found tracing the perpendicular line from the exterior point. So, to construct the angle bisector we have followed these steps:

1. Construct the hyperbolic ray that passes through the two first given points, C and D.
2. Construct the hyperbolic ray that passes through the two last given points, D and E.
3. Plot the tangent line, r, to the ray that joins C and D and that contains D. To plot the tangent line, we use that the tangent from a point in a circumference is the perpendicular line to the normal one, which is given by the radius. Thus, we plot perpendicular line to the radius that goes through point D.
4. Draw, as in step (3), the tangent line, s, to the ray that joins D and E and contains D. We want to plot the Euclidean bisector of the Euclidean angle, since we are working in a conformal model. We have to decide which two of the four Euclidian rays are those we have to consider since we have had four Euclidean angles. To make it we use the fact exposed in the first paragraph.

5. Plot the perpendicular line to the tangent that contains C.
6. Consider the intersection of the former line with r.
7. Plot the Euclidean ray that starts at D and passes through the former intersection. This ray will form the Euclidean angle.
8. Plot the perpendicular line to s that contains E.
9. Consider the intersection of the former line with s.
10. Plot the Euclidean ray that starts at D and passes through the former intersection. This ray is the other one that will define the Euclidean angle.
11. Plot the Euclidean angle bisector from the point in step (6), point D and the point in step (9). From the Euclidean angle bisector we will plot the hyperbolic angle bisector. It should be fulfilled that the Euclidean angle bisector is tangent to the hyperbolic one. With this we construct the hyperbolic angle bisector in the following way:
12. Plot the perpendicular line to the Euclidean angle bisector that contains D.
Consider the intersection between this line and the boundary line.
13. Plot the circumference with center the intersection in step (12) and that passes through D. The hyperbolic angle bisector  we are searching for is on this hyperbolic line. It only remains to consider three points of the arc that has to give us the hyperbolic angle bisector. The initial point is D. The end point is one of the two intersection points of the circumference with the boundary line.
14. Plot the Euclidean segment that joins the intersection points in steps (6) and (9).
15. Consider the intersection point of the segment of the former step with the Euclidean angle bisector. This intersection exists because the segment is formed with a point of each ray that defines the angle we want to bisect.
16. Plot the perpendicular line to the Euclidean angle bisector that passes through the former point.
17. Consider the intersection between the line in step (16) and the circumference we have constructed in step (14). The first intersection point will be considered as the intermediate point. We consider as the first point the first found when we go through the line from the point in step (16).
18. To choose the final point, construct the arc of circumference that passes through D, the point in step (17) and the antipodal point of D (which we can find tracing the line that passes through the center of the circumference and D). The intersection point of this arc of circumference with the boundary line is the final point.
19. Plot the arc with origin in D, passing through the point in step (17) and with final point, the point in step (18). This gives us the hyperbolic angle bisector. To see that the hyperbolic bisector will always be well-constructed it is necessary to prove that the step (18) will always be well defined, that is, the perpendicular line will always intersect with the circumference.
We have already commented during the construction the existence of the other intersections. Observe that the steps (7) and (10) are not necessary for the construction even though they can be enlighting. In them we construct the rays that define the Euclidian angle even though this tool only needs three points (a point of a ray, the vertex and a point of the other ray). So, with the intersection points defined in the steps (6) and (9) we could make the construction.

List of tools
Hyperbolic geometry