Hyperbolic perpendicular line

To plot the perpendicular for a point we will distinguish two cases: the point belongs to the line or the point is exterior to the line.

Point in the line

In Absolute Geometry it is known that can be plotted one and only one perpendicular line to a line that goes through a given point. To construct the hyperbolic perpendicular we will use a similar way to the one that we used for the perpendicular bisector.

To plot the hyperbolic perpendicular we need a hyperbolic line and a point that belongs to this straight line. To assure that the point belongs to the line, first we mark two points, to construct the straight line and then we mark a point in the constructed line.
  1. Construct the hyperbolic line that goes through the two given points.
  2. Mark the point, P, of the line where we want to plot the perpendicular.
  3. Plot the tangent line to the plotted hyperbolic line in the step (1) that passes through the point P. To construct this line we can plot the normal straight line, from the radius and to consider its perpendicular.
  4. Plot the circumference that passes through P and has center in the intersection point between the tangent line and the boundary line.
  5. Consider the two intersection points between this circumference and the boundary line.
  6. Plot the arc with origin in one of the point of former intersection, passing through P and ending in the other intersection point. This is the perpendicular line that we wanted to construct.


To prove that this hyperbolic line we have constructed is the perpendicular line, it is enough seeing that it forms right angles with the initial line. As in the case of the bisector angle, we have that this is true because we can find an inversion with center in the boundary line that assures us that the angles are right angles. The inversion which it is necessary to consider is the same circumference which we search to prove that the constructed line is the perpendicular line. This circumference has the center, O, in the boundary line. To prove it, we can consider any line going through the center of the inversion and intersect with the given hyperbolic line in two points, C and D. The power from the point O in the hyperbolic line allows us to affirm that: OP·OP=OC·OD and to be OP the radius of the circumference of inversion we obtain that points C and D are inverse. So, the angle formed by the circumference of inversion and the segment PC transforms into the angle formed by the same circumference and the segment PD. As these two angles are adjacent and equal they are, by definition, right angles. From here we can affirm that the plotted hyperbolic line is the perpendicular.

It is necessary to remark that on using this tool, it will not ask us for the third point, but it will put one point by default. To obtain the perpendicular from the point that we want, we have only to move the point that the program has given us.
Exterior point
To construct the perpendicular in the case that the point is exterior to the line we will use the hyperbolic bisector and the parallel lines. We know that given a hyperbolic line and an exterior point exists one and only one perpendicular line to the given line that goes through the point. This affirmation is true since it can be proved in the Absolute Geometry. To be able to use this tool, we have to give three points. The two first points will determine the line and the third point we will suppose that it is the exterior point. In this construction we will suppose that the hyperbolic line is a semicircumference with center in the boundary line.
To construct the perpendicular line for an exterior point we will follow these steps:
  1. Construct the hyperbolic line that goes through the two first points.
  2. Construct the two parallel lines to the hyperbolic line just constructed and that go through the third point.
  3. Follow the steps that allow us to construct the angle bisector but only until the step (14). We have constructed the circumference that contains the bisector. 
  4. Plot the arc of circumference that will determine the perpendicular line. Consider the arc with origin in one of the two intersection points between the former circumference and the boundary line, going through P and finishing in the other  intersection point.


In this way we can determine the perpendicular from an exterior point. All steps are well defined, if we suppose that the two given points do not belong to a perpendicular line to the boundary line, since they use constructions which we already know that, in this situation, we can always make. The plotted line is perpendicular because we have constructed it from the parallel lines and we know that given an exterior point, if we plot the perpendicular for this point and the parallel lines we have that the angles that each one of the parallel lines forms with the perpendicular are equal. So, if we have the two parallel lines and another line that divides the angle that they form, this line will be the perpendicular line since it will fulfill the same properties as the perpendicular line and we know that it is unique.

List of tools
Hyperbolic geometry