Girolamo Saccheri was a Jesuit priest living from 1667 to 1733. Before he died he published a book entitled Euclides ab omni nævo vindicatus ( Euclid Freed of Every Flaw). It sat unnoticed for over a century and a half until rediscovered by the Italian mathematician Beltrami.
He wished to prove Euclid's Fifth Postulate from the other axioms. To do so he decided to use a reductio ad absurdum argument. He assumed the negation of the Parallel Postulate and tried to arrive at a contradiction. He studied a family of quadrilaterals that have come to be called Saccheri quadrilaterals . Let S be a convex quadrilateral in which two adjacent angles are right angles. The segment joining these two vertices is called the base . The side opposite the base is the summit and the other two sides are called the sides . If the sides are congruent to one another then this is called a Saccheri quadrilateral. The angles containing the summit are called the summit angles .
Theorem 11.1: In a Saccheri quadrilateral

Proof: Let M be the midpoint of AB and let N be the midpoint of CD.

and
, so that by SAS
, which implies that
. Also, since
then
we may apply the SSS criterion to see that
. Then, it is clear that
.
is perpendicular to both
and
. Now
,
, and
. Thus by SAS
. This
means then that
. Also,
and
. By
SSS
and it follows that
. They are supplementary angles, hence they must be right angles.
Therefore
is perpendicular to
.
Using the analogous proof and triangles
and
,
we can show that
is perpendicular to
.
Saccheri considered all possible cases of such a quadrilateral. They are:
is a right angle | HRA, the hypothesis of the right angle. |
is an obtuse angle | HOA, the hypothesis of the obtuse angle. |
is an acute angle | HAA, the hypothesis of the acute angle. |
We shall see that HRA is equivalent to Euclid's Postulate V, so we may take HOA or HAA as negations of Postulate V. The Three Musketeers Theorem implies that if one of HRA, HOA, or HAA holds for one quadrilateral, then it holds for all.
Theorem 11.2: In a Saccheri quadrilateral
on the base AB under the assumption HRA, HOA, or HAA we have
, AB>CD, or AB<CD, respectively, and the angle sum of a triangle is equal to, greater than, or less than two right angles, respectively.
Proof: Let M and N denote the midpoints of AB andCD, respectively. We will work with the assumption HAA, since the others are similar or already known. We wish to show that AB<CD.
Suppose CD<AB, then CN<BM. There is a point
so that
and
. Thus,
must be greater than a right angle, since
is between
and
.
is a Saccheri quadrilateral on the base NM, which implies that
, which implies that
is greater than a right angle. In
is an exterior angle and hence is greater than
. Recall that by hypothesis,
is an acute angle. Thus, we have an acute angle greater than an angle that is greater than a right angle, our contradiction. Therefore, AB<CD.
Now, suppose we have a right triangle
with right angle at B. Construct AD perpendicular to AB, with
. Assuming HAA, CD>AB and
since the greater side subtends the greater angle. Now,
is a right angle. Thus, the angle sum of
is less than two right angles.
Consider two parallel lines k and
, and let
and
. Let
and construct lines perpendicular to
through X and Y. Call these lines
and
.

We are interested in the angles
and
. If both angles are acute, then the two lines
and
have a common perpendicular between the segments UX and VY. If one of the angles is a right angle and the other is acute, then the two lines already have a common perpendicular. Suppose that
is acute and
is obtuse. If we move V away from U along
then
may change to a right angle or remain obtuse.
Theorem 11.3: As above and assuming HAA, if as V moves away from U along
,
remains obtuse, then
is asymptotic to
.
Saccheri now shows the following:
Theorem 11.4: Given a point P not on a line
, there are three classes of lines through P:
,
, and
and hence asymptotic to
.
Saccheri now concludes ``the hypothesis of the acute angle is absolutely false; because it is repugnant to the nature of straight lines.''
The flaw in Saccheri's work was observed by the Swiss mathematician J.H. Lambert in 1786. He himself contributed to the work in non-euclidean geometries with the following.
A convex quadrilateral three of whose angles are right angles is called a Lambert quadrilateral .
Under the HAA assumption the following are true.
Theorem 11.5: The fourth angle of a Lambert quadrilateral is acute.
Theorem 11.6: The side adjacent to the acute angle of a Lambert quadrilateral is greater than its opposite side.
Theorem 11.7: In a Saccheri quadrilateral the summit is greater than the base and the sides are greater than the altitude.
Proof: Using Theorem 12.1 if M is the midpoint of AB and N
is the midpoint of CD, then
is a Lambert quadrilateral. Thus,
AB>MN and, since
, both sides are greater than the altitude.
Also, applying Theorem 12.1 DN>AM. Since
and
it follows that CD>AB, so that the summit is greater than the base.