By Poonen B.
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Extra info for Elliptic curves (introduction)
96) and 3. We now take advantage of the equality of the angles (see Eqs. 93)) and consider the segments AO1 , BO2 , and CO3 which intersect the sides BC, AC, and AB at the points A2 , B2 , and C2 , respectively (see Fig. 43). We apply the theorem of bisectors to the triangles AO1 A1 , BO2 B1 , and CO3 C1 and use Eq. 94) to obtain Eqs. 98) 54 3 Fundamentals on Geometric Transformations Fig. 43 Another translation of the ratios (Sect. 7) and OC1 O3 C1 O3 C2 . 99) Using Eqs. 99), we have: AO1 BO2 CO3 + + AA2 BB2 CC2 = AA2 + A2 O1 BB2 + B2 O2 CC2 + C2 O3 + + AA2 BB2 CC2 = AA2 BB2 CC2 O1 A2 O2 B2 O3 C2 + + + + + AA2 BB2 CC2 AA2 BB2 CC2 =3+ OA1 OB1 OC1 + + AA1 BB1 CC1 = 3 + 1 = 4.
The final position of a shape, when it is a result of several translations, is independent of the order in which the translations take place. 1 Let ABCD be a quadrilateral with AD = BC and let M, N be the midpoints of AB and CD, respectively. Show that MN is parallel to the bisector of the straight semilines AD, BC. Proof We translate the sides AD and BC to the positions MM 1 and MM 2 , respectively (see Fig. 8). It is sufficient to show that M1 MN = NMM 2 . 8) 24 3 Fundamentals on Geometric Transformations Fig.
19 Homothety (Sect. 5) 3. A characteristic criterion of homothety. A necessary and sufficient condition for a shape S to be homothetic to a shape S with ratio r = 0, 1 is that for each pair −−→ − → of points A, B of S there is a pair of points A , B of S such that A B = r AB. Proof Let S be homothetic to S with respect to O with ratio r = 0, 1. Let A, B be points of S and let A , B be their corresponding homologous points on S . 26) −→ − → OB = r OB. 29) −−→ − → A B = r AB. 30) Therefore, that is, or equivalently, 32 3 Fundamentals on Geometric Transformations Fig.