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By David Gans

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Stated in words, the defect of a triangle is equal to the sum of the defects of any two triangles into which it is subdivided. Consequently, if triangle BCD is subdivided into two triangles with defects d'2 and d2, then d2=d'2 + d"2. Substituting this in (1) we obtain the equation d = d^ + d2 + d2, relating the defect of the original triangle ABC to the defects of the three triangles into which it has been subdivided. Clearly, if any of these three triangles is subdivided into two triangles and this process of subdivision is continued further, then at each stage the defect of triangle ABC equals the sum of the defects of the triangles into which triangle ABC has been subdivided.

The proof of the following converse of Theorem 37 is left as an exercise. Theorem 38. If two lines have a common perpendicular, there are transversals which cut the lines so as to form equal alternate interior angles (or equal corresponding angles). The only transversals with this property are those which go through the point on that perpendicular which is midway between the lines. 7. PARALLELS WITH A COMMON PERPENDICULAR 51 We have already learned something about the distance between two parallel lines.

This proves statement (j). EXERCISES 1. Using only the basis E, show that if the angle-sum of a triangle is always less than 180°, then the fourth angle in a Lambert quadrilateral is always acute. 2. It can be shown on the basis E that if the angle-sum is less than 180° in one triangle, then it is less than 180° in all triangles. Using this fact, verify that in Legendre's proof the defect of triangle AEF (Fig. II, 9) is actually more than 2a. 30 8. II. ATTEMPTS TO PROVE THE FIFTH POSTULATE THE DISCOVERY OF NON-EUCLIDEAN GEOMETRY Well written as it was, Legendre's 1833 publication viewed from our time was out of date even before it appeared, so greatly had some of his contemporaries progressed with the problem of Postulate 5.

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