Gaps in Euclid’s Arguments 19 A A A A B B B B C C C C D D D D M M ` ` ` O O O O F F F G G G (a) (b) (c) (d) Fig. 1.14. The “proof” that every triangle has two equal sides. AOC have two pairs of equal angles and share the common side AO, so it follows from Proposition I.26 that the sides AB and AC are equal. In all of the remaining cases, we assume that the extension of AD is not perpendicular to BC . Let BC be bisected at M (Proposition I.10), let ` be the line perpendicular to BC at M (Proposition I.11), and let AD be extended if necessary (Postulate 2) so that it meets ` at O. There are now three possible cases, depending on the location of O. CASE 1: O lies inside triangle ABC (Fig. 1.14(b)). Draw BO and CO (Postulate 1). Note that the triangles BMO and CMO have two pairs of equal corresponding sides (MO is common and BM D CM ), and the included angles BMO and CMO are both right, so the remaining sides BO and CO are also equal by Proposition I.4. Now draw OF perpendicular to AB and OG perpendicular to AC (Proposition I.12). Then triangles AFO and AGO have two pairs of equal corresponding angles and share the side AO, so Proposition I.26 implies that their remaining pairs of corresponding sides are equal: AF D AG and FO D GO. Now we can conclude that BFO and CGO are both right triangles in which the hypotenuses BO and CO are equal and the legs FO and GO are also equal. Therefore, the Pythagorean theorem (Proposition I.47) together with Common

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