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By David Gans

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Sample text

30). Since the angles 48 III. PARALLELS WITH A COMMON PERPENDICULAR Fig. Ill, 7 at C and D are acute, AD > EF in quadrilateral AEFD and BC > EF in quadrilateral EBCF(§3, Theo. 33). This proves the second part of the theorem. On noting that these quadrilaterals can also be viewed so that AE, DF and EB, FC are arms, we again use Theorem 33, obtaining and DF>AE FC > EB. Combining these inequalities gives DF+FC>AE + EB, or DC > AB, which completes the proof of the theorem. Lines AB, CD in the preceding discussion are parallel since they have a common perpendicular, and AD, EF are two distances from line CD to line AB that are unequal.

The distance from a point on either parallel to the other increases as the point recedes from the perpendicular in either direction. Proof. Let g, h (Fig. Ill, 10) be parallels with a common perpendicular which meets them in points E, F. On h choose any point B other than F and let C be its projection on g. Since ECBF is a Lambert quadrilateral, < 1 is acute and BC > FE (§6, Theo. 35). Thus the distance from h to g is less F J Ί B 2 1 B' 3 1 1 c C Fig. Ill, 10 when measured along the common perpendicular than along any other perpendicular from h to g.

Using only the basis E, show how to construct a quadrilateral with three right angles. 2. Using only the basis E, show that any Saccheri quadrilateral can be subdivided so as to give two congruent Lambert quadrilaterals. Hence show that each of Lambert's hypotheses implies the same-named Saccheri hypothesis. 3. Using only the basis E, show that from any given Lambert quadrilateral a certain Saccheri quadrilateral can be determined. Hence show that each Saccheri hypothesis implies the same-named Lambert hypothesis.