Thesauros Edebiyat mathematics Στοιχεῖα

Στοιχεῖα

Στοιχεῖα Euclid

Kitap 12

17 If then we conceive straight lines joined from the points P, S, Q, T, R, U to A, there will be constructed a certain polyhedral solid figure between the circumferences BO, KO, consisting of pyramids of which the quadrilaterals KBPS, SPQT, TQRU and the triangle URO are the bases and the point A the vertex.
17 And, if we make the same construction in the case of each of the sides KL, LM, ME as in the case of BK, and further in the case of the remaining three quadrants, there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids, of which the said quadrilaterals and the triangle URO, and the others corresponding to them, are the bases and the point A the vertex.
17 I say that the said polyhedron will not touch the lesser sphere at the surface on which the circle FGH is.
17 Let AX be drawn from the point A perpendicular to the plane of the quadrilateral KBPS, and let it meet the plane at the point X; [XI. 11] let XB, XK be joined.
17 Then, since AX is at right angles to the plane of the quadrilateral KBPS, therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. [XI. Def. 3]
17 Therefore AX is at right angles to each of the straight lines BX, XK.
17 And, since AB is equal to AK, the square on AB is also equal to the square on AK.
17 And the squares on AX, XB are equal to the square on AB, for the angle at X is right; [I. 47] and the squares on AX, XK are equal to the square on AK. [id.]
17 Therefore the squares on AX, XB are equal to the squares on AX, XK.
17 Let the square on AX be subtracted from each; therefore the remainder, the square on BX, is equal to the remainder, the square on XK; therefore BX is equal to XK.
17 Similarly we can prove that the straight lines joined from X to P, S are equal to each of the straight lines BX, XK.
17 Therefore the circle described with centre X and distance one of the straight lines XB, XK will pass through P, S also, and KBPS will be a quadrilateral in a circle.

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