Question

Prove that no two lines in hyperbolic geometry are equidistant from one another by showing that the distance from one line to another cannot have the same value in more than two places. Please prove geometrically, not algebraically.

Answer 1

To prove that no two lines in hyperbolic geometry are equidistant from one another , first we require some prerequisites , so let us consider the following quadrilateral as shown below :

                                      

We have the following theorem for above figure as : The line joining the midpoints of the upper and lower bases of the Saccheriquadrilateral (called the altitude) is perpendicular to both. Therefore, the upper base andlower base lie on parallel lines sharing a common perpendicular.

First we need to prove this theorem 1 :

PROOF :

• Let E and F be the midpoints of the lower base and upper base respectively. Then let anglesα,α′,β,β′,γ,γ′,μandμ′be as in Figure 1 ABOVE.
• DEA∼=CEB by SAS (SIDE - ANGLE - SIDE )
• ThusDE∼=CE,α=α′, and β=β′.
• Now from SSS ( SIDE - SIDE - SIDE ) we have , it follows that DEF∼=CEF.Hence γ=γ′, and since these angles are supplementary, each must be 90degree. Also,μ=μ′,and so α+μ=α′+μ= 90◦, again since these angles are supplementary.
• ThusEF⊥AB and EF⊥CD, and it follows from that AB‖CD with common perpendicular EF.
• The proof of the following theorem uses the fact that a triangle cannot have two angles summing to more than two right angles.

Consider an other theorem as follows : Consider a quadrilateral with a lower base that makes right angles with its two arms.

(i) If the upper base angles are unequal, so are the arms.

(ii) If the arms are unequal, so are the upper base angles, with the greater upper baseangle opposite the greater arm.

PROOF of theorem 2 : consider the following figure below :

                                      

• The upper base angles of the Saccheri quadrilateral are equal . This is proved theorem or fact. for (i) case.
• For (ii), suppose that BC >AD in quadrilateral ABCD. Let E be the point on the segment BC such that AD∼=BE. Then ABED is a Saccheri quadrilateral, so∠ADE∼=∠BED by the theorem of upper base angles are equal. Note that ∠ADC= ∠ADE+ ∠EDC.
• Now we know that in a triangle, an exterior angle is greater than either of the interior and opposite angles. So we have∠BED=∠ADE >∠ECD,and therefore∠ECD <∠ADE+∠EDC=∠ADC
• This means that they hold in both Euclidean and hyperbolic geometry .
• n order to examine some results that hold in hyperbolic geometry but not Euclidean geometry we must first state a replacement for the Parallel Postulate.
• Hence The upper base angles of the Saccheri quadri-lateral are acute.

Consider theorem 3 : In the Saccheri quadrilateral:

(i) the altitude is shorter than the arms, and

(ii) the upper base base is longer than the lower base.

PROOF : Consider the figure below :

                                        

• On the Saccherri quadrilateral ABCD, let Eand Fbe the midpoints of the lowerbase and upper base respectively.
• According to theorem 1 above we have , that EFis perpendicular to both the upper and lower base. By the Hyperbolic Parallel Postulate,∠C and ∠D are acute.
• Now consider theorem 2 above we have , AD >EFin AEFDand CB >EFin EBCF.
• To prove (ii), consider EFDA as having lower base EF and arms AE and DF. Theorem 2 gives us that DF >AE and similarly FC >EB. ThereforeDF+FC >AE+EB or DC >AB. Thus the upper base is longer than the lower base.
• Let us consider the Saccheri quadrilateral from the previous construction .

                                 

In Theorem 2 we proved that lines DC and AB are parallel with common perpendicular EF.

• Theorem 3 gives us that AD >EF and BC >EF. Thus the parallel lines DC and AB are not equidistant.
• Hence PROVED.