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10). Then (ACD) + (DCB) = (ACE) + (ECB) = 2 right angles. 17 When two lines intersect, the opposite angles are equal. 16 we have (see Fig. 11) (BEA) + (AEC) = 2 right angles = (AEC) + (CED). 4, one can subtract the angle (BEA) = (CED). 18 In a triangle, an external angle at a vertex is greater than each internal angle at another vertex. 52 3 Euclid’s Elements Fig. 11 Fig. 12 Proof Given the triangle ABC as in Fig. 12, it is claimed that (ACD) > (ABC), (ACD) > (BAC). 12, let E be the middle point of the segment AC.

When it has been useful to make the presentation more “accessible”, we sometimes combine several results in a single statement. We make an exception for Book 1, which we present in more detail, following rather closely the original text of Euclid: our purpose is to illustrate rather faithfully Euclid’s axiomatics and method of deduction. We also present in some detail the main arithmetical results of Book 7, for further use in the Appendices of this book. 1 1. 2. 4. 8. A point is that of which there is no part.

11 Bisect an angle. 5, draw the equilateral triangle DEF (see Fig. 7). 4, we get the equality (ADF ) = (AEF ). 7 to the triangles ADF and AEF , we obtain that AF bisects the angle (BAC). 12 Bisect a segment. 11, draw the equilateral triangle ACB on the given segment AB and the bisector of the angle ACB, which cuts AB at D (see Fig. 8). 7 to the triangles ACD and BCD forces the conclusion. 50 3 Euclid’s Elements Fig. 7 Fig. 13 Draw a perpendicular at a given point of a line. Proof We refer to Fig.

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