Section 8.3 Saccheri quadrilaterals
¶The key idea in associating the defect of a triangle with its area is to use Saccheri quadrilaterals, which are quadrilaterals with two right angles, and two congruent legs, as shown in Figure 8.3.1. The base of the Saccheri quadrilateral is the side between the right angles, and the opposite side is called the summit. It is easy to show that the remaining angles, called summit angles, are congruent by SASAS.
Any triangle has an associated Saccheri triangle, obtained by connecting the midpoints of two sides, and dropping perpendicular line segments from the vertices, as illustrated in Figure 8.3.2. 1 Before proceeding, construct the associated Saccheri quadrilateral by selecting both checkboxes in the figure. It is easy to see that \(\triangle AMP\cong\triangle BMQ\) and \(\triangle CNR\cong\triangle BNQ\) by AAS (which follows from SAS). Thus, all three perpendicular line segments, namely \(AP\text{,}\) \(BQ\text{,}\) and \(CR\text{,}\) are congruent, which is enough to ensure that the quadrilateral \(ACRP\) is a Saccheri quadrilateral. Explicitly, there are right angles at \(P\) and \(R\text{,}\) and the two legs \(AP\) and \(CR\) are congruent.
Why does the associated Saccheri quadrilateral matter? First of all, it's clear from Figure 8.3.2 that the area of a triangle is the same as that of its associated Saccheri quadrilateral. Because there are two pairs of congruent triangles, as discussed above, all we've done in constructing the latter is to move two triangles into other positions. Furthermore, the defect of a triangle is the same as that of its associated Saccheri quadrilateral! Don't forget that the defect of a quadrilateral is obtained by subtracting the angle sum from \(360^\circ\text{.}\) The angles of the triangle have just been redistributed into the quadrilateral – and two right angles have been added. Just right for the defects to agree.
The next piece of the puzzle is that the summit angles of the Saccheri quadrilateral (at \(A\) and \(C\) in Figure 8.3.2) are congruent, by SASAS congruence of the quadrilateral with its mirror image. But this means that the summit angles tell us the defect! More precisely, the defect of the quadrilateral is \(D=360-(180+2s)\text{,}\) where \(s\) is the measure of each summit angle, so \(s=\frac12(180-D)\text{.}\)
Finally, two hyperbolic Saccheri quadrilaterals with the same summit and summit angles (the side and angles opposite the right angles) must be congruent. This assertion can be proved using the same reasoning as in the proof given in Section 8.2 of AAA congruence. If the legs of the two Saccheri quadraliterals are not congruent, then one fits inside the other, resulting in a rectangular piece between them with defect zero. But rectangles do not exist in hyperbolic geometry.
Thus, hyperbolic Saccheri quadrilaterals are completely determined by specifying the length of their summit and either summit angle (and hence both). We will use this property in Section 8.4 to show that hyperbolic triangles with the same defect have the same area.