Archimedis Opera cum Apollonii Conicorum Libri IIII

than the excess X. And then by drawing the straight lines A C, A E, A V; B D, B E, B V; c it will be as before, the conical surface C V A + segment C A symbol: greater than triangle CVA. And the conical surface A V E + segment A E symbol: greater than triangle A V E; and thus combined, the conical surface C V E + segments C A, A E symbol: greater than triangle CVA + AVE; symbol: greater than triangle C V E. By similar reasoning, the conical surface D V E + segments D B, B E symbol: greater than triangle D V E; and combined, the conical surface C A V B D + segments C A, A E, D B, B E symbol: greater than triangle C V E + D V E d = triangle C V D + X. Whence since segments C A + A E + D B + B E symbol: less than X, the conical surface C A V B D will be symbol: greater than the triangle C V D. Q.E.D.

Thus, ἀκριβείας precision/exactness, Archimedes demonstrates a matter that is sufficiently clear in itself, just as the three following ones are no less αὐτοφανεῖς self-evident and αὐτοπίςους self-credible/self-authenticating: indeed, he is averse to multiplying axioms and postulates beyond necessity.

Prop. XIII.

Fig. 16.

If straight lines (AC, BC) are drawn tangent to the circle (ADB) which is the base of the cone (VADB), existing in the same plane as the circle, and meeting one another; and from the contacts (A, B) and from the meeting point (C) straight lines (AV, BV, CV) are drawn to the vertex of the cone (V); the triangles (AVC, BVC) enclosed by the tangents and the lines joined to the vertex of the cone are greater than the conical surface taken away by them.

Let the arc A B be bisected at D, and through D let a tangent E F be drawn, and let V E, V F be connected; a 20. 1. and E C + F C symbol: greater than E F; wherefore by adding the common A E + B F, it will be A C + B C symbol: greater than A E + E F + B F. b 1. 6. Consequently triangle A V C + B V C symbol: greater than triangle A V E + B V F + E V F (since the height of these triangles is common). Let the excess be X, not less than the segments A E D, B F D: now because the pyramidal surface E A V B F, whose base is the trapezium E A B F, c 4. ax. huj is greater than the included conical surface A V B, with the segment A D B (the perimeter of the triangle A V B being the common limit); and by subtracting the common segment A D B, the triangles A V E, E V F, B V F with the segments A E D, B F D are greater than the conical surface A V B D: therefore, all the more are the triangles A V E, E V F, B V F with X greater than the same d hyp. surface; d that is, triangle A V C + B V C symbol: greater than conical surface AVBD.
If X should be less than the segments A E D, B F D, let the arcs A D, B D be bisected, and their halves, and so continuously e 9 hujus. until the remaining segments A L G, G K G, D M H, H N B have turned out smaller than X. And by drawing the straight lines V L, V K, V M, V N, the demonstration will proceed similarly as in the previous one.