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similar and lequiangular (6. 6.); and of GN, is the duplicate (20. 6.) ratio therefore the angle A EB is equal to of that which BM has to GN; and the angle FLG: But AEB is equal the ratio of the polygon ABCDE to (21. 3.) to AMB, because they stand the polygon FGHKL is the duplicate upon the same circumference; and the angle FLG is, for the same reason, equal to the angle FNG: There
E fore also the angle AMB is equal to FNG: And the right angle BAM is equal to the right (31. 3.) angle GFN;
M wherefore the remaining angles in the
к triangles ABM, FGN are equal, and they are equiangular to one another : (20.6.) of that which BA has to CF: Therefore as BM to GN, so (4.6.) Therefore as the square of BM to the is BA to GF; and therefore the due square of GN, so is the polygon plicate ratio of BM to GN is the same ABCDE to the polygon FGHKL. (10. def. 5. & 22. 5.) with the dupli- Wherefore similar polygons, &c. cate ratio of BA to ĠF: But the ra. Q. E. D. tio of the square of BM to the sqnare
PROP. II. THEOREM.
Circles are to one another as the squares of their diameters. Let ABCD, EFGH be two circles, the square EFGH is greater than and BD, FH their diameters: As the half of the circle. Divide the circumsquare of BD to the square of FH, so ferences EF, FG, GH, HE, each is the circle ABCD, to the circle into two equal parts in the points K, EFGH.
L, M, N, and join EK, KF, FL, LG, For, if it be not so, the square of GM, MH, HN, NE: Therefore each BD shall be to the square of FH, as of the triangles EKF, FLG, GMH, the circle ABCD is to some space HNE is greater than half of the segeither less than the circle EFGH, or ment of the circle it stands in ; begreater than it.* First, let it be to a cause, if straight lines touching the space S less than the circle EFGH; circle be drawn through the points and in the circle EFGH describe the K, L, M, N, and parallelograms upsquare EFGH: This square is greater on the straight lines EF, FG, GH, than half of the circle EFGH ; be- HE, be completed; each of the tricause if, through the points E, F, G, angles EKF, FLG,GMH, HNE shall H, there be drawn tangents to the be the half (41. 1.) of the parallelocircle, the square EFGH is half (41. gram in which it is: But every seg1.) of the square described about the ment is less than the parallelogram circle; and the circle is less than the in which it is: Wherefore each square described about it; therefore of the triangles EKF, FLG, GMH,
• For there is some square equal to the circle ABCD; let P be the side of it : and to three straight lines BD, FH, and P, there can be a fourth proportional ; let this be Q. Therefore the squares of these four straight lines are proportionals; that is, to the squares of BD, FH, and the circle ÅBCD, it is possible there may be a fourth proportional. Let this be S. And in like manner are to be understood some things in some of the following propositions.
HNE is greater than half the segment greater (14. 5.) than the polygon of the circle which contains it: And EKFLGMHN: But it is likewise if these circumferences before-named less, as has been demonstrated ; be divided each into two equal parts, which is impossible. Therefore the and their extremities be joined by square of BD is not to the square of straight lines, by continuing to do FH, as the circle ABCD is to any
space less than the circle EPGH. In X
the same manner, it may be demonк N
strated, that neither is the square of B
FH to the square of BD, as the cir.
cle EFGH is to any space less than PL M
the circle ABCD. Nor is the square G
of BD to the square of FH, as the this, there will at length remain seg. circle ABCD is to any space greater ments of the circle, which, together, than the circle EFGH : For, if posshall be less than the excess of the sible, let it be so to T, a space greater circle EFGH, above the space S: than the circle EFGH: Therefore, Because, by the preceding Lemma, if inversely, as the square of FH to the from the greater of two unequal magsquare of BD, so is the space T to nitudes there be taken more than its the circle ABCD. But as the space half, and from the remainder more than its half, and so on, there shall X
N at length remain a magnitude less than the least of the proposed mag
H nitudes. Let then the segments EK,
I M KF, FL, LG, GM, MH, HN, NE be those that remain and are together less than the excess of the circle EFGH, above S: Therefore the rest of the circle, viz. the polygon
T EKFLGMHN, is greater than the space S. Describe likewise in the cir. T is to the circle ABCD, so is the cle ABCD the polygon AXBOCPDR circle EFGH to some space, which similar to the polygon EKFLGMHN: must be less (14. 5.) than the circle As therefore, the square of BD is to ABCD, because the space T is greater, the square of FH, so (1. 12.) is the by hypothesis, than the circle EFGH. polygon AXBOCPDR to the poly- Therefore as the square of FH is to gon EKFLGMHN; But the square the square of BD, so is the circle of BD is also to the square of FH, as EFGH to a space less than the circle the circle ABCD is to the space S: ABCD, which has been demonstrated Therefore as the circle ABCD is to to be impossible: Therefore the the space S, so is (11.5.) the poly, square of BD is not to the square of gon AXBOCPDR to the polygon FH as the circle ABCD is to any EKFLGMHN: But the circle ABCD space greater than the circle EFGH: is greater than the polygon contained And it has been demonstrated, that in it: wherefore the space s, is neither is the square of BD to the
For, as in the foregoing note at, 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.
square of FH, as the circle ABCD to the square of FH, so is the circ! any space less than the circle EFGH: ABCD to the circle EFGH. Circles Wherefore as the square of BD to therefore are, &c. Q. E. D.
PROP. III. THEOREM.
Every pyramid having a triangular base, may be divided into two equal
and similar pyramids having triangular bases, and which are similar to the whole pyramid ; and into two equal prisms, which together are greater than half of the whole pyramid.
Let there be a pyramid of which not in the same plane with them, they the base is the triangle ABC and its contain equal (10. 11.) angles; theresertex the point D: The pyramid fore the angle EHG is equal to the ABCD may be divided into two equal angle KDL. Again, because EH, and similar pyramids D HG are equal to KD, DL, each to having triangular bases,
each, and the angle EHG equal to the and similar to the whole;
angle KDL; therefore the base EG and into two equal
is equal to the base KL: And the prisms which together K L triangle EHG equal (4. 1.) and simi. are greater than half of
lar to the triangle KDL: For the same the whole pyramid.
reason, the triangle AEG is also eDivide AB, BC, CA,
qual and similar to the triangle HKL. AD, DB, DC, each into
Therefore the pyramid, of which the
с two equal parts in the
base is the triangle AEG, and of points E, F, G, H, K, L, and join which the vertex is the point H, is EH, EG, GH, HK, KL, LH, EK, equal (C. 11 ) and si- D KF, FG. Because AE is equal to milar to the pyramid EB, and AH to HD, HE is parallel the base of which is the (2. 6.) to DB : For the same reason, triangle KHL, and ver, HK 18 parallel to AB: Therefore tex the point Di And K HEBK is a parallelogram, and HK because HK is parallel equal (34. I. to EB: But EB is equal to AB a side of the trito AE; therefore also AE is equal to angle ADB, the trianHK : And AH is equal to HD; gle ADB is equianguwherefore EA, AH are equal to KH, lar to the triangle HDK, HD, each to each; and the angle and their sides are proportionals: (4. EAH is equal (29. 1.) to the angle 6.) Therefore the triangle ADB is si. KHD; therefore the base EH is equal milar to the triangle HDK. And for to the base KD, and the tria:gle AEH the same reason, the triangle DBC equal (1. 1.) and similar to the trian- is similar to the triangle DKL: and gle HKD: For the same reason, the the triangle ADC to the triangle triangle AGH is equal and similar to HDL; and also the triangle ABC to the triangle HLD: And because the the triangle AEG: But the triangle two straight lines EH, HG, which AEG is similar to the triangle HKL, meet one another, are parallel to KD, as before was proved ; therefore the DL that meet one another, and are triangle ABC is similar (21. 6.) to the
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.
triangle HKL. And the pyramid of and the vertices the points H, D: bewhich the base is the triangle ABC, cause, if EF be joined, the prism har. and vertex the point D), is therefore ing the parallelogram EBFG for its similar (B. 11. & 11. def. 11.) to the base, and KH the straight line oppo. pyramid of which the base is the tri- site to it, is greater than the pyramid angle HKL, and vertex the same of which the base is the triangle EBF, point D: But the pyramid of which and vertex the point K; but this pythe base is the triangle HKL, and ramid is equal (C. 11.) to the pyramid vertex the point D), is similar, as has the base of which is the triangle AEG, been proved, to the pyramid the base and vertex the point H: because they of which is the triangle AEG, and are contained by equal and similar vertex the point H: Wherefore the planes: Wherefore the prism having pyramid, the base of which is the tri- the parallelogram EBFG for its base, angle ABC, and vertex the point D, and opposite side KH, is greater than is similar to the pyramid of which the the pyramid of which tbe base is the base is the triangle AEG and vertex triangle AEG, and vertex the point H: Therefore each of the pyramids H: And the prism of which the base AEGH, HKLD is similar to the is the parallelogram EBFG, and opwhole pyramid ABCD: And because posite side KH is eqnal to the prism BF is equal to FC, the parallelogram having the triangle GFC for its base, EBFG is double (41. 1.) of the trian- and HKL the triangle opposite to it; gle GFC: But when there are two and the pyramid of which the base is prisms of the same altitude, of which the triangle AEG, and vertex H, is one has a parallelogram for its base, equal to the pyramid of which the and the other a triangle that is half base is the triangle HKL, and vertex of the parallelogram, these prisms D: Therefore the two prisms before are equal (40. 11.) to one another; mentioned are greater than the two therefore the prism having the paral- pyramids of which the bases are the lelogram CBFG for its base, and the triangles AEG, HKL, and vertices straight line KH opposite to it, is e- the points H, D. Therefore the whols qual to the prism having the triangle pyramid of which the base is the tri GFC for its base, and the triangle angle ABC, and vertes the point D, HKL opposite to it; for they are is divided into two equal pyramids siof the same altitude, because they milar to one another, and to the whole are between the parallel (15.11.) planes pyramid ; and into two equal prisms; ABC, HKL: And it is manifest that and the two prisms are together each of these prisms is greater than greater than half of the whole pyraeither of the pyramids of which the mid. Q. E. D. triangles AEG, HKL are the bases,
PROP. IV. THEOREM. If there be two pyramids of the same altitude, upon triangular bases, and
each of lkem be divided into two equal pyramids similar to the whole pyramid, and also into two equal prisms; and if each of these pyra. mids be divided in the same manner as the first two, and so on : As the base of one of the first two pyramids is to the base of the other, so shall all the prisms in one of them be to all the prisms in the other that are produced by the same number of divisions.
Let there be two pyramids of the bases, ABC, DEF, and having their same altitude upon the triangular vertices in the points G, H; and let
each of them be divided into two equal in the pyramid ABCG are equa. ta pyramids similar to the whole, and one another, and also the two prisms into two equal prisms; and let each in the pyramid DEFH equal to one of the pyramids thus made be con- 'another, as the prism of which the ceived to be divided in the like man. base is the parallelogram KBXL and ner, and so on : As the base ABC is opposite side MO, to the prison having to the base DEF, so are all the prisms the triangle LXC for its base, and in the pyramid ABCG to all the prisms OMN the triangle opposite to it; so in the pyramid DEFH made by the is the prism of which the base (7. 5.) same number of divisions.
is the parallelogram PEVR, and op : Make the same construction as in posite side TS, to the prism of which the foregoing proposition: And be the base is the triangle RVF, and opcause BX is equal to XC, and AL to posite triangle STY. Therefore, comLC, therefore XL is parallel (2. 6.) to ponendo, as the prisms KBXLMO, AB, and the triangle ABC similar to LXCOMN together are unto the the triangle LXC: For the same reason, the triangle DEF is similar to RVF: And becaase BC is double of CX, and EF double of FV, therefore BC is to CX, as EF to FV: And upon
Y BC, CX are described the similar and
T Y similarly situated rectilineal figures ABC, LXC; and upon EF, FV, in like manner, are described the similar figures DEF, RVF: Therefore, as
1 the triangle ABC is to the triangle prism LXCOMN: so are the prisms LXC, so (22. 6.) is the triangle DEF PEVRTS, RVFSTY to the prism to the triangle RVF, and, by permu- RVFSTY: And permutando, as the tation, as the triangle ABC to the tri- priems KBXLMO, LXCOMN are angle DEF, so is the triangle LXC to to the prisms PEVRTS, RVFSTY; the triangle RVF: And because the so is the prism LXCOMN to the planes ABC, OMN, as also the planes prism RVISTY: But as the prism DEF, STY, are parallel, the per- LXCOMX to the prism RVFSTY, pendiculars drawn from the points so is, as has been proved, the base G, H to the bases ABC, DEF, which ABC to the base DEF: Therefore, by the hypothesis, are equal to one
as the base ABC to the base DEF, another, shall be cul each into two so are the two prisms in the pyramid equal (17. 11.), parts by the planes ABCG to the two prisms in the pyraOMN, STY, because the straight mid DEFH: And likewise if the pylines GC, HF are cut into two equal ramids now made, for example, the parts in the points N, Y by the same two OMNG, STYH be divided in planes: Therefore the prisms LXC- the same manner; as the base OMN OMN, RVFSTY are of the same al is to the base STY, so shall the two titude ; and therefore 'as the base prisis in the pyramid OMNG be to LXC to the base RVF; that is, as the two prisms in the pyramid STYH: the triangle ABC to the triangle DEF, But the base OMN is to the base STY so (Cor. 32. 11.) is the prism having as the base ABC to the base DEF: the triangle LXC for its base, and therefore, as the base ABC to the OMN the triangle opposite to it, to base DEF, so are the two prisms in the prism of which the base is the tri. the pyramid ABCG to the two prisma angle RVF, and the opposite triangle in the pyramid DEFH; and 80 are STY: And because the two prisms the two prisms in the pyramid OMNG