# Substitute for Bolyai-Lobachevsky Axiom

In euclidean geometry, the axiom "Sum of inner angles of a triangle equals $\pi$" is a well known substitute for the parallel postulate. I wonder if the analogue occurs in hyperbolic geometry. Is "Sum of inner angles of a triangle less than $\pi$" a substitute for Bolyai-Lobachevsky Axiom? If so, I would like to know how can I prove Bolyai-Lobachevsky Axiom from it. I already proved the converse.

Thanks.

Bolyai-Lobachevsky Axiom: Given a line $r$ and a point $P \notin r$, there are at least two lines $s,t$ through $P$ not intersecting $r$.

• What do you take as the Bolyai-Lobachevsky axiom ? – Rene Schipperus Nov 11 '16 at 4:27
• Given a line $r$ and a point $P \notin r$, there are at least two lines $s,t$ through $P$ not intersecting $r$. – user286485 Nov 11 '16 at 17:01

After some puzzeling I found a quite simple solution:

Note that this solution only proofs the existence of two (diverging) parallel lines through a point not on a line , but that is all what is needed :)

Start with a $\triangle ABC$

Then construct a congruent $\triangle BCD$ $d(CD) = d(AB), d(BD)= d(AC) , m(\angle BCD) = m (\angle ABC), m(\angle BDC) = m (\angle BAC)$

and a congruent $\triangle ACE$ $d(CE) = d(AB), d(AE)= d(BC) m(\angle ECA) = m (\angle ABC), m(\angle AEC) = m (\angle BAC)$

From this follows:

$m(\angle ECD ) = m(\angle ECA ) +m(\angle ACB ) +m(\angle BCD) = m(\angle ABC ) +m(\angle ACB ) +m(\angle BAC) \not= \text{ 2 right angles}$ Therefore $E$, $C$ and $D$ are not colliniear and $EC$ and $CD$ are different lines

The line $BC$ makes equal angles with $CD$ and $AB$ , $m(\angle ABC) = m(\angle BCD)$ by Euclid 1.27 the line $CD$ and the line $AB$ do not intersect

The line $AC$ makes equal angles with lines $CE$ and $AB$ , $m(\angle BAC) = m(\angle ACE)$ by Euclid 1.27 the line $CE$ and the line $AB$ do not intersect

So both lines $EC$ and the line $CD$ do not intersect the line $AB$

We therefore have two lines through a point $(C)$ not intersecting a line $(AB)$