[b]a) [/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 is exterior to the segment YZ as shown in the top figure (and on page 279 of textbook). [br]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. [br]Through AAS triangle congruence, we see that:[br][math]\triangle ADX\cong\triangle BDY[/math] and [math]\triangle AXE\cong\triangle CZE[/math]. [br]From the first triangle congruence, we know that [math]AX\cong BY[/math]. From the second, we know that [math]AX\cong CZ[/math]. Thus, [math]CZ\cong BY[/math]. Thus, we can see that quadrilateral ZYBC is a Saccheri quadrilateral because we defined the angles at Z and Y to be perpendiculars and [math]CZ\cong BY[/math]. [br]Then[br][math]\angle A+\angle B+\angle C=\angle CAB+\angle ABC+\angle BCA[br][/math][br]   [math]=\angle XAD-\angle EAX+\angle DBC+\angle BCZ+\angle ZCE[/math][br]   [math]=\angle YBD+\angle DBC+\angle BCZ[/math][br]   [math]=\angle YBC+\angle BCZ[/math][br]   [math]=2\cdot\angle BCZ[/math]. [br]Because quadrilateral ZYBC is a Saccheri quadrilateral, [math]\angle BCZ[/math] is acute. Thus, [math]\angle A+\angle B+\angle C=2\cdot\angle BCZ<2\cdot90^\circ<180^\circ[/math]. [br][br][b]b)[/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 [math]\triangle ADE\cong\triangle CDE\cong\triangle BDY[/math]. Thus, the quadrilateral EYBC (or ZYBC) is a Saccheri quadrilateral because the angles at E and Y are right angles and [math]EC\cong YB[/math]. [br]Then[br][math]\angle A+\angle B+\angle C=\angle CAB+\angle ABC+\angle BCA[/math][br]   [math]=\angle YBD+\angle YBC-\angle YBD+\angle BCD+\angle DCE[/math][br]   [math]=\angle YBC-\angle ECB[/math][br]   [math]=0<180^\circ[/math].