Proof 11.10
In the hyperbolic plane, the sum of the angles in any triangle is less than 180 degrees.
a) Suppose that ABC is a triangle in the hyperbolic plane. Let D be the midpoint of the segment AB and let E be the midpoint of the segment AC. Construct segments AX, BX, and CZ that are perpendicular to the line DE, with the points X,Y,Z lying on the line DE. Consider the case when point X is exterior to the segment YZ as shown in the top figure (and on page 279 of textbook).
Note, this figure shows A to the left of YZ. Without loss of generality, if A was to the right, our argument would still hold.
Through AAS triangle congruence, we see that:
and .
From the first triangle congruence, we know that . From the second, we know that . Thus, . Thus, we can see that quadrilateral ZYBC is a Saccheri quadrilateral because we defined the angles at Z and Y to be perpendiculars and .
Then
.
Because quadrilateral ZYBC is a Saccheri quadrilateral, is acute. Thus, .
b) Suppose that ABC is a triangle in the hyperbolic plane. Let D be the midpoint of the segment AB and let E be the midpoint of the segment AC. Construct segments AX, BX, and CZ that are perpendicular to the line DE, with the points X,Y,Z lying on the line DE. Consider the case when point X equals point Z (and thus also equals point E). Then we see that . Thus, the quadrilateral EYBC (or ZYBC) is a Saccheri quadrilateral because the angles at E and Y are right angles and .
Then
.