Archimedis Opera cum Apollonii Conicorum Libri IIII (1675) - Page 28

Prop. XIV.

Fig. 17.

If there are two straight lines (A C, B D) on the surface of a right cylinder (A C D B), the surface of the cylinder (A C F D B F A) intercepted by the straight lines is greater than the parallelogram (A C D B) enclosed by the straight lines (A C, B D) on the surface of the cylinder, and those (A B, C D) which join their endpoints.

a 20. I.
b 1. 6.

Let the arcs A B, C D be bisected at E, F; and let A E, B E, C F, D F be drawn: and because A E + E B a > A B, b it will be parallelogram A E F C + B E F D > parallelogram A B D C (the height of all being equal): let X be the excess not less, first, than the segments A E, B E, C F, D F; now the cylindrical surface A E B D F C + segments A E B, C F D c > parallelogram A E F C, B E F D + triangles A E B, C F D (the parallelogram A B C D being the common limit). Therefore, by subtracting the common triangle A E B + C F D, the cylindrical surface A E B D F C + segments A E, B E, C F, D F > parallelogram A E F C + B E F D d = parallelogram A B C D + X. Wherefore, since X is equal to or less than these segments, it is clear that the cylindrical surface A E B D F C is greater than the parallelogram A B C D.

c 4 ax. hujus.

d hyp.

e 8 hujus.

If X should be less than these segments, let the arcs A E, B E, C F, D F be bisected, and their halves, e until the remaining segments A G, G E, E H, H B; C L, L F, F M, M D are smaller than X: then by drawing the straight lines, as in the figure, it will be (as before) { parallelogram A G L C + G E F L > parallelogram A E F C. { parallelogram B H M D + H E F M > parallelogram B E F D. And because the Cylindrical surface A E B D F C + segments A E B, C F D > parallelogram A G L C, G E F L, B H M D, H E F M + rectilinear figure A G E H B, C L F M D (the parallelogram A B D C being the common limit), therefore, having subtracted these common rectilinear figures, the cylindrical surface A E B D F C + segments A G, G E, E H etc. > parallelogram A G L C, G E F L, B H M D, H E F M > parallelogram A E F C + B E F D = parallelogram A B D C + X. Whence, since X is greater than these segments, it is clearly evident that the cylindrical surface A E B D F C is greater than the parallelogram A B C D. Q.E.D.

const.

Prop. XV.

Fig 18.

If there are two straight lines (A C, B D) on the surface of a certain right cylinder, and from the ends of the lines certain lines (A E, B E & C F, D F) are drawn tangent to the circles which are the bases of the cylinder, existing in the same plane, and meeting; the parallelograms (A E F C, B E F D) enclosed by the subtangents and the sides of the cylinder will be greater than the surface of the cylinder, intercepted by the straight lines (A C, B D) which are on the surface of the cylinder.