# saccheri quadrilateral - how does base=summit violate hyperbolic parallel axiom?

I drew diagonals across the quadrilateral and was able to prove that the summit angles are right angles by SSS and CPCTC. Therefore the two congruent triangles creats a quadrilateral with an angle sum of 360 degrees. So each triangle must equal 180 degrees which violates the theorem that the angle sum of triangles in hyperbolic planes are less than 180 degrees.

I just can't seem to tie all this together with how the summit=base violates the hyperbolic parallel axiom which is defined as "given a line A and a point B off the line, there is more than one line through B that does not meet A" in my textbook.

Any help is appreciated! Thank you!

In hyperbolic geometry the base is shorter than the summit

as help:

Turn the whole Saccheri quadrilateral on its side.

The sides are diverging parallel lines (the original base is part of the common perpendicular) and is the shortest segment between them. assuming that base = summit is the same as assuming the parallel postulate.

Good luck

A proof sketch that base = summit cannot exist in hyperbolic geometry:

Basic theorems needed (to be proved seperately, and valid in both hyperbolic and euclidean geometry)

For every two lines $x$ and $y$ that intersect at a point $Q$:

$I_1$: Part of $y$ is one halfplane bounded by $x$ part of $y$ is the other halfplane bounded by $x$.

$I_2$: on each halfplane bounded by $x$ here is a point $Y$ on $y$ that is a given distance $j$ from $x$

$NI$: For every two lines $x$ and $y$ that do not intersect the whole of $x$ is on one halfplane bounded by $y$

Hyperbolic paralel axiom:

$HP$ "given a line $a$ and a point $B$ off the line, there is more than one line through $B$ that does not meet $a$"

Then lets call the line of which the summit is a part $l$ and the line of which the base is a part $m$

1 The lines $l$ and $m$ by the construction you made are equidistant lets call the distance between them $i$

2 by $NI$ the line $m$ is on one and only one halfplane bounded by $l$

3 All points a distance $i$ from $l$ on the side of $m$ are on $m$ (needs proof)

4 on $l$ choose a random point $P$

5 For every other line $n$ also trough $P \not= l$

6 by $I_2$ there will be a points $N$ on the side of $m$ of $l$ that is a distance $i$ from $l$

7 then by 2 point $N$ is on $m$

8 therefore $n$ and $m$ intersect. (both go trough point $N$)

9 therefore there is no second line that does not intersect $m$.

10 this is opposite of what the Hyperbolic parallel axiom says, so the proposition that base = summit is impossible

This is just a rough proof sketch , maybe it is difficult to fill in the gaps and maybe there are easier proofs (I hope so)