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By David Gans

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Hence BG = AF and AF = CH, so that BG = CH. Thus BCHG is a Saccheri quadrilateral with summit BC. The triangle ABC we have used is one in which angles ADE and AED are acute, but the same result would be obtained if one of these angles is obtuse (Fig. Ill, 23) or if one is a right angle (Fig. Ill, 24). 64 III. PARALLELS WITH A COMMON PERPENDICULAR Fig. Ill, 23 D = F=G Fig.

The summit angle KLN of this quadrilateral, which is acute by the hyperbolic parallel postulate, is the fourth angle of the Lambert quadrilateral JKLM. To show that LK > MJ and ML > JK one proceeds just as in the proof of Theorem 34. EXERCISES 1. Prove that the fourth angle of a Lambert quadrilateral is acute by using an argument involving the negation of a substitute for Postulate 5. 2. Show that a quadrilateral whose opposite sides are equal is a parallelogram, but that the converse is not true.

6. Show that if two noncongruent Saccheri quadrilaterals have equal bases, the one with the longer arms has the smaller summit angles. 9. THE DEFECT OF A TRIANGLE 59 7. Prove that two Saccheri quadrilaterals are congruent* if the summit and summit angles of one are equal, respectively, to the summit and summit angles of the other. 8. Prove that two Lambert quadrilaterals are congruent* if the acute angle and an adjacent side in one are equal to the acute angle and an adjacent side in the other.