1. A non-Euclidean model in the unit disk
A non-Euclidean model in the unit disk, a set of Euclidean theorems
A non-Euclidean model is presented in this paper. Main goal of
paper is to show that how non-Euclidean model in the unit disk
can lead us to a set of Euclidean theorems.

2. A non-Euclidean model in the unit disk
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Abstract
A non-Euclidean model is presented in this paper. Main goal of paper is to show that how non-Euclidean model
in the unit disk can lead us to a set of Euclidean theorems.
Keywords
Hyperbolic geometry, Poincare disk model, parallel angle, unit disk
Introduction
The Poincare disk model is a geometric model that represents the hyperbolic geometry in the unit disk. At first
Poincare defined his model in the unit disk as follows:
1- Each point inside the unit circle is an ordinary point of model.
2- Each point on the unit circle is an ideal point of model.
3- Each line (Poincare-line or p-line) of model is part of a circle that has two properties
3-a) cut the unit circle perpendicular.
3-b) is inside the unit circle. (See fig.1)
psi
D
B
A
C
F
E
G
H
K
90°
Fig.1- Three p-lines GH, GK and EF and also a psi-line CD are shown

3. A non-Euclidean model in the unit disk
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Fig.2- Two p-lines and angle between them at their intersection point
4- Angle between two p-lines is equal to angle between two circles (that represent lines) at their
intersection point.(See fig.2)
5- Distance between two ordinary points A and B is defined as logarithm of cross ratio as follows:
AB= Ln (AB, FE) 1
Where E and F are ideal points of a p-line that passes through points A and B
Then Poincare proved that the hyperbolic geometry and his model are equivalent geometrically. In other words
if a theorem is valid in his model then it is valid in the hyperbolic geometry and vice versa.
Now let develop the Poincare disk model via change cut angle between p-line and the unit circle. Assume that
ψ-model is a model that their lines are ψ-lines. A ψ-line (psi-line in fig.1) cut the unit circle at angle of ψ. Now
you may define a ψ-model as follows:
-Each point inside the unit circle is an ordinary point of model.
-Each point on the unit circle is an ideal point of model.
-Each line (ψ-line) of model is part of a circle so that has two properties:
a) It cut the unit circle at angle of ψ.
b) It is inside the unit circle.

4. A non-Euclidean model in the unit disk
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-Angle between two ψ-lines is equal to angle between two circles (that represent lines) at their
intersection points.
-Distance between two ordinary points A and B is defined as logarithm of cross ratio as follows:
AB= Ln (AB, CD) 1
C and D are ideal points of a ψ-line that passes through points A and B (See fig.1)
Simple calculation yields:
(AB, CD) =
1+tan 𝜓.tan (𝑑 cos 𝜓)
1−tan 𝜓.tan (𝑑 cos 𝜓)
2
“d” is Euclidean distance of AB when A is located at the center of the unit disk. (See fig.1)
Let concentrate on a special model named zero-model. It means that all lines of zero-model are tangent to the
unit disk internally. Note that all lines of zero-model cut the unit disk at just one point. We are interested in
theorems that should be proved in this beautiful model.
First we accept that in the unit disk:
- For any two ordinary points A, B there is a unique parallel angle that may be defined as in the Poincare
model.
- For any four ordinary points A, B, C and D we can say that:
AB=CD if and only if parallel angle (See fig.3) of AB is equal to parallel angle of CD
psi
B
A
B
A S
psi
S
Fig.3- Definition of parallel angle at point B (Right angle at A, parallel angle at B, AS and BS are boundary-
parallel)
A number of theorems in the zero-model are listed as follows (without proof):

5. A non-Euclidean model in the unit disk
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1- Theorem 1: Parallel angle of two ordinary points is unique. (Note that four parallel angles exist,
because in fig.3 one may draw AS and BS so that right angle is at B and parallel angle at A. Also two
lines may pass through points A and B)
2- Theorem 2: Zero-rectangle exists.
3- Theorem 3: Two zero-lines that have one ideal point in common are equi-distance.
4- Theorem 4: One Euclidean circle passes through vertices of a zero-rectangle.
5- Theorem 5: There are two ideal points Q, T for any two ordinary points A, B (See fig.5) so that
absolute[cross ratio of (AB,QT)]=1.
6- Theorem 6: Given a zero-rectangle (ABCD). (See fig.5). So we have four ideal points (S, Q, R and T).
Absolute of cross ratio of these ideal points is equal to 1.
7- Theorem 7: Each zero-circle is Euclidean circle. These circles have distinct centers.
8- Theorem 8: Triangle inequality is valid. (Note that distance between two ordinary points is defined
according to the equation-1)
9- Given three ψ-lines, find the unit circle. (See fig.4) This is the Apollonius-problem in its general form.
All of these theorems should be proved geometrically. They are really right!!!
C1
C3
C2
W1
W3
W2
CC
Fig.4- Apollonius problem in general form may be translated in language of ψ-model.
These theorems should be translated in language of Euclidean geometry.

6. A non-Euclidean model in the unit disk
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Theorem 1: Two ellipses A and B have a foci “O” in common. Also assume that their constant value is equal.
These A and B ellipses cut each other in points “P” and “Q” and have distinct foci A and B respectively. Select
arbitrary point M on the line of PQ and draw lines from it to points A and B until cut the ellipses A and B at M
and N respectively. If OM and ON cut ellipses A and B at points X and Y respectively. Prove that three lines AX,
BY and PQ have one point in common. It is sufficient that prove the problem for Poincare half-plane. (See fig.5)
Some theorems mentioned above are valid in any ψ-model for example theorem 8. Pure geometric proof for
theorem 8 may be very beautiful since I have not found it yet!
Problem 9: Apollonius-problem: Given three circles (C1, C2, C3 in fig.4), find a circle (C in fig.4) that cut
them at angle of ψ.
A simple solution for problem-9 is to find a conformal mapping (for example Mobius mapping) so that three
contact points of a circle (that is tangent to all three given circles) are mapped in form of right-hand figure
symmetrically. So the right-hand unit disk is obtained. Inverse Mobius mapping of right-hand unit disk gives the
left-hand unit disk.
D
C
BA
S
Q
R
T
Fig.5- Apollonius problem in general form may be translated in language of ψ-model.