inversely, as the square of FH to the square of BD, so is the space T to the circle ABCD; but as the space IT is to the circle ABCD, so is the circle EFGH to some space, which must be less * * 14.5. than the circle ABCD, because the space T is greater, by hypothesis, than the circle EFGH; therefore as the square of FH is to the square of BD, so is the circle EFGH to a space less than the circle ABCD, which has been demonstrated to be impossible; therefore the square of BD is not to the square of FH as the circle ABCD is to any space greater than the circle EFGH: and it has been demonstrated, that neither is the square of BD to the square of FH, as the circle ABCD to any space less than the circle EFGH: wherefore, as the square of BD to the square of FH, so is the circle ABCD to the circle EFGHS Circles, therefore, are, &c. Q. E. D. # For as, in the foregoing note, it was explained how it was possible there could be a fourth proportional to the squares of BD, FH, and the circle ABCD, which was named S; so, in like manner, there can be a fourth proportional to this other space, named T, and the circles ABCD, EFGH. And the like is to be understood in some of the following propositions. Because, as a fourth proportional to the squares of BD, FH, and the circle ABCD, is possible, and that it can neither be less nor greater than the circle EFGH, it must be equal to it. PROP. III. THEOR. divided into two equal and similar pyramids Let there be a pyramid of which the base is the triangle ABC and its vertex the point D: the pyramid ABCD may be divided into two equal and similar pyramids D having triangular bases, and similar to the whole; and into two equal prisms which together shall be greater than half of the whole K L pyramid. Divide AB, BC, CA, AD, DB, DC, each into two equal parts in the points E, F, G, H, K, L, and B join EH, EG, GH, HK, KL, LH, EK, KF, FG. Because AE is equal to EB, * 2.6. and AH to HD, HE is parallel* to DB: for the same reason, HK is parallel to AB: therefore *34. 1. HEBK is a parallelogram, and HK equal* to EB: but EB is equal to AE ; therefore also AE is equal to HK: and AH is equal to HD; wherefore EA, AH are equal to KH, HD, * 29. 1. each to each; and the angle EAH is equal* to the angle KHD; therefore the base EH is equal to the base KD, and the triangle AEH *4.1. equal * and similar to the triangle HKD. For the same reason, the triangle AGH is equal and similar to the triangle HLD. Again, because the two straight lines EH, HG, which meet one another, are parallel to KD, DL, that meet one' another, and are not in the same plane with *10.11. them, they contain equal * angles; therefore the angle EHG is equal to the angle KDL; and because EH, HG are equal to KD, DL, each to each, * * * are* and the angle EHG equal to the angle KDL ; therefore the base EG is equal to the base KL, and the triangle EHG equal * and similar to * 4. 1. the triangle KDL. For the same reason, the triangle AEG is also equal and similar to the triangle HKL. Therefore the pyramid of which the base is the triangle AEG, and of which the vertex is the point H, is equal * and *.C.11. similar to the pyramid, the base of which is the triangle KHL, and vertex the point D. And because HK is D parallel to AB, a side of the triangle ADB, the triangle ADB is equiangular to the triangle HDK, and their sides K L proportionals: *4.6. therefore the triangle ADB is similar to the triangle HDK: and G for the same reason, the triangle DBC is similar to the triangle B DKL; and the triangle ADC to the triangle HDL; and also the triangle ABC to the triangle AEG; but the triangle AEG is similar to the triangle HKL, as before was proved; therefore the triangle ABC is similar* to the tri- * 21. 5. angle HKL: and therefore the pyramid of which the base is the triangle ABC, and vertex the point D, is similar* to the pyramid of which the *B.11.& base is the triangle HKL, and vertex the same point D: but the pyramid of which the base is the triangle HKL, and vertex the point D, is similar, as has been proved, to the pyramid the base of which is the triangle AEG, and vertex the point H; wherefore the pyramid, the base of which is the triangle ABC, and vertex the point D, is similar to the pyramid of which the base is the triangle AEG, and vertex H: therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD. And because BF is equal to FC, the parallelogram EBFG is double * of * 41. 1. the triangle GFC: but when there are two prisms of the same altitude, of which one has a parallelo : 11. Def. 11. D; * gram for its base, and the other a triangle that is, * 40. 11. half of the parallelogram, these prisms are equal* to one another; therefore the prism having the parallelogram EBFG for its base, and thestraight line KH opposite to it, is equal to the prism having the triangle GFC for its base, and the triangle HKL opposite to it ; for they are of the same altitude, because they are between the 15.11. parallel * planes ABC, HKL: and it is manifest that each of these prisms is greater than either of the pyramids of which the triangles AEG, HKL are the bases, and the vertices the points H, because, if EF be joined, the prism having the parallelogram EBFG for its base, and KH the straight line opposite to it, is greater than the pyramid of which the base is the triangle EBF, and vertex the point K: but this pyramid is *C. 11. equal* to the pyramid, the base of which is the triangle AEG, and vertex the point H; because they are contained by equal and similar planes: wherefore the prism having the parallelogram EBFG for its base, and opposite side KH, is greater than the pyramid of which the base is the triangle AEG, and vertex the point H: and the prism of which the base is the parallelogram EBFG, and opposite side KH, is equal to the prism having the triangle GFC for its base, and HKL the triangle opposite to it; and the pyramid of which the base is the triangle AEG, and vertex H, is equal to the pyramid of which the base is the triangle HKL, and vertex D: therefore the two prisms before mentioned are greater than the two pyramids of which the bases are the triangles AEG, HKL, and vertices the points H, D. Therefore the whole pyramid of which the base is the triangle ABC, and vertex the point D, is divided into two equal pyramids sirnilar to one another, and to the whole pyramid; and into two equal prisms; and the two prisms are together greater than half of the whole pyramid. Q.E.D PROP. IV. THEOR. upon triangular bases, and each of them bé Let there be two pyramids of the same altitude upon the triangular bases ABC, DEF, and having their vertices in the points G, H; and let each of them be divided into two equal pyramids similar to the whole, and into two equal prisms; and let each of the pyramids thus made be conceived to be divided in the like manner, and so on : as the base ABC is to the base DEF, so shall all the prisms in the pyramid ABCG be to all the prisms in the pyramid DEFH made by the same number of divisions. Make the same construction as in the foregoing proposition: and because BX is equal to XC, and AL to LC, therefore XL is parallel * * 2.6. to AB, and the triangle ABC similar to the triangle LXC. For the same reason, the triangle DĚF is similar to RVF. And because BC is double of CX, and EF double of FV, therefore t +0.5. BC is to CX, as EF to FV: and upon BC, CX are described the similar and similarly situated rectilineal figures ABC, LXC; and upon EF, FV, in like manner, are described the similar figures DEF, RVF : therefore, as the triangle ABC is to the triangle LXC, so * is the triangle * 22.6 DEF to the triangle RVF, and, by permutation, as the triangle ABC to the triangle DEF, so is the triangle LXC to the triangle RVF. And * сс |