Transport of segments

In the Euclidean Geometry, we know that there are some applications that preserve the distances and the angles. These applications, called isometries, are the rotations and the translations. The isometries bring lines to lines and segments to segments of the same length. To apply a translation we have to know the direction and the distance and to apply a rotation we have to know the angle.

In the Hyperbolic Geometry we have also isometries. In the half-plane model, the isometries are composition of inversions with respect to circumferences with center in the boundary line. To fix an isometry it is necessary to fix a point in the boundary line (which will be the center of the circumference of inversion) and a distance (which will be the radius of the circumference of inversion).

From the isometries we will be able to transport the objects of the Hyperbolic Geometry.
This sketch allows us to transport a segment to another position. We have to fix the beginning of the transported segment and the direction, that is, it is necessary to fix a point and a ray. To make the transport we will use two isometries.

To make the transport of a segment CD to the ray with end E we have followed these steps:

  1. Plot the hyperbolic segment CE.
  2. Plot the hyperbolic perpendicular bisector of the hyperbolic segment CE.

  3. We consider as inversion circumference the circumference that describes this perpendicular bisector. We apply the inversion to the points C and D. C transforms into E since, by construction, the inversion circumference passes through the midpoint of the segment CE.

  4. Plot the Euclidean line that joins the points C and E.
  5. Consider the intersection point between the former line and the boundary line. This point, P, is the center of the inversion circumference.
  6. Measure the radius, r, of the circumference of inversion. We can make it selecting the circumference and the tool "Measure/Radius".
  7. Calculate the distance between the point D and its image by the inversion that we are considering. We can find this distance from the inversion's definition, that is, from the formula: PD = r·r/PC.
  8. Plot the Euclidean circumference of center P and radius the former result.
  9. Plot the Euclidean line that joins the points P and D.
  10. Consider the intersection between the circumference in step (7) with the line in step (8). This point, F, is the image of D by the inversion.
  11. Plot the hyperbolic bisector of the angle that has for vertex the point E and for side EF and the given ray.

  12. We consider as another inversion circumference, the circumference that determine the bisector. We apply the inversion in
    the points E and F. The point E is fixed for the inversion to belong to the circumference of inversion. As the inversions preserve the angles, the ray EF transforms with the given ray. Thus, the image of F by this inversion will determine the end of the transported segment. From now, we cannot follow a method analogous to that of the steps (3)-(9) because we know the image of a fixed point, so, we cannot plot the line for the point and its image.

  13. Plot the segment that joins the point E to the intersection point of the bisector with the boundary line.
  14. Consider the Euclidean midpoint of the former segment.
  15. Plot the perpendicular line to the segment that goes through the midpoint.
  16. Consider the intersection point of the former line with the boundary line. This point is the center of the circumference of inversion since it has been constructed from the perpendicular bisector of a segment that joins two of the points of the circumference.
  17. Plot the Euclidean line that joins the former point with F.
  18. Consider the intersection, J, of this line with the given ray. This is the image of F and the end of the segment that we searched.
  19. Plot the hyperbolic segment that has endpoints in E and J. This is the transported segment of the segment CD.

List of tools
Hyperbolic geometry