The elements of non-Euclidean plane geometry and trigonometry, by H. S. Carslaw.

54

NON-EUCLIDEAN GEOMETRY

[CH. III.

Thus the quadrilateral ACPQ has the angles at C, P, and Q
right angles.
Therefore, by ~ 30, the angle at A, namely / CAD, must be
acute.
It follows that the sum of the angles of any right-angled
triangle must be less than two right angles.
Case II. Consider now any triangle, not right-angled.
Every triangle can be divided into two right-angled triangles
by drawing the perpendicular from at
least one angular point to the opposite A
side (Fig. 32).
Let AD be the perpendicular referred
to in the triangle ABC, and let the
angles e', a", /3, y be as in the figure.
Then A + B +C=(o + f) +(o" + y).
But e' +/3 < 1 right angle y
and ac" + y < 1 right angle. B D C
Therefore A + B + C < 2 right angles. FIG- 32.
It should be noticed that no use has been made of the
Postulate of Archimedes in proving this result.
The difference between two right angles and the sum of the
angles of a triangle will be called the Defect of the Triangle.
COROLLARY. There cannot be two triangles with their angles
equal each to each, which are not congruent.
It is easy to show that if two such triangles did exist, we
could obtain a quadrilateral with the sum of its angles equal to
four right angles. We have simply to cut off from one of the
triangles a part congruent with the other. But the sum of
the angles of a quadrilateral cannot be four right angles,
if the sum of the angles of every triangle is less than two
right angles.
~32. Not-intersecting Lines.
It follows from the Theorem of the External Angle (I. 16)
that if two straight lines have a common perpendicular, they
cannot intersect each other. And they cannot be parallel,
since this would contradict Hilbert's Axiom of Parallels [cf.
~ 26 (3)].