Sidebilder
PDF
ePub

that EG taken from DE is not greater than its half; but BH Book XII. taken from AB is greater than its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and GF is not greater than the half of GD, but HK is greater than the half of HA, therefore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C. Q. E. D.

And if only the halves be taken away, the same thing may, in the same way, be demonstrated.

PROP. I. THEOR.

Similar polygons inscribed in circles are to one another as the squares of their diameters.

Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHKL; and let BM, GN, be the diameters of the circles: As the square of BM is to the square of GN, so is the polygon ABCDE to the polygon FGHKL.

Join BE, AM, GL, FN: And because the polygon ABCDE is similar to the polygon FGHKL, and similar polygons are divided into similar triangles; the triangles ABE, FGL, are

[merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small]

similar and equiangular b; and therefore the angle AEB is b 6. 6. equal to the angle FLG: But AEB is equal to AMB, 21. 3. because they stand upon the same circumference: and the angle FLG is, for the same reason, equal to the angle FNG: Therefore also the angle AMB is equal to FNG: And the right angle BAM is equal to the right d angle GEN: Where- d 31. 3. fore the remaining angles in the triangles ABM, FGN are

Book XII. equal, and they are equiangular to one another: Therefore

as BM to GN, soe is BA to GF; and therefore the duplie 4. 6.

cate ratio of BM to GN is the samef with the duplicate f 10.def.5.

ratio of BA to GF: But the ratio of the square of BM to g 20. 6. the square of GN is the duplicate 8 ratio of that which BM

& 22. 5.

[blocks in formation]

has to GN: and the ratio of the polygon ABCDE to the polygon FGHKL is the duplicate 5 of that which BA has to CF: Therefore as the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore, a similar polygons," &c. Q. E. D.

PROP. II. THEOR.

Sce N.

Circles are to one another as the squares of their diameters.

Let ABCD, EFGH be two circles, and BD, FH their diameters: As the square of BD to the square of FH, so is the circle ABCD to the circle EFGH.

For, if it be not so, the square of BD shall be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it*. First, let it be to a space S less than the circle EFGI: and in the circle EFGH describe the square EFGH: This square is greater than half the circle EFGH; because if, through the points E, F, G, H, there be drawn tangents to the circle, the square

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 ABCD, 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.

EFGH is half a the square described about the circle; and Book XII. the circle is less than the square described about it: therefore the square EFGH is greater than half the circle. Divide a 41. 1. the circumferences EF, FG, GH, HE each into two equal parts, in the points K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE: Therefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle it stands in; because if straight lines touching the circle be drawn through the points K, L, M, N, and parallelograms upon the straight lines EF, FG, GH, HE be completed; each of the triangles EKF, FLG, GMH, HNE shall be the halfa of the parallelogram in which it is: But every segment is less than the parallelogram in which it is: Wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which contains it: And if these circumferences before named be divided each into two equal parts, and their extremities be joined by straight lines, by continuing

[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small]

to do this, there will at length remain segments of the circle, which, together, shall be less than the excess of the circle EFGH, above the space S: Because, by the preceding Lemma, if from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes. Let then the segments EK, 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 EKFLGMHN, is greater than the space S. Describe likewise in the circle ABCD the polygon AXBOCPDR, similar to the polygon EKFLGMHN: As therefore the square of BD is to the square of FH, sob is the polygon b 1. 12. AXBOCPDR to the polygon EKFLGMHN: But the

Book XII. square of BD is also to the square of FH, as the circle ABC1)

is to the space S: Therefore, as the circle ABCD is to the c 11. 5. space S, so is the polygon AXBOCPDR to the polygon

EKFLGMHN: But the circle ABCD is greater than the d 14. 5. polygon contained in it; wherefore the space S is greater a

than the polygon EKFLGMHN: But it is likewise less, as has been demonstrated; which is impossible. Therefore the square of BD is not to the square of FH, as the circle ABCD is to any space less than the circle EFGH. In the same manner, it may be demonstrated, that neither is the square of FH, to the square of BD, as the circle EFGH is to any space less than the circle ABCD. Nor is the square of BD to the square of FH, as the circle ABCD is to any space greater than the circle EFGH: For, if possible, let it be so to T, a space greater than the circle EFGH: Therefore, inversely, as the square of FH to the square of BD, so is the space T

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

to the circle ABCD. But as the space + T is to the circle ABCD, so is the circle EFGH to soine space, which must be less d than the circle ABCD, because the space T is greater,

+ For, as in the preceding 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 A BCD, EFGH. And the like is to be understood in some of the following propositions.

by hypothesis, than the circle EFGH. Therefore, as the Book XIT. square of FH is to the square of BD, so is the circle EFGH Sa 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 EFGH. “ Circles, therefore, are," &c. Q. E. D.

PROP. III. THEOR.

Every pyramid having a triangular base, may be See N. 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 the whole pyramid.

Let there be a pyramid of which the base is the triangle A BC, and its vertex the point D: The pyramid ABCD may be divided into two equal and similar pyramids having triangular bases, and 1 similar to the whole; and into two equal prisms which together are greater than half the whole pyramid. Divide AB, BC, CA, AD, DB, DC,

H
each into two equal parts, in the points
E, F, G, H, K, L, and join EH, EG,

K L.
GH, HK,KL, LH, EK, KF, FG. Be-
cause AE is equal to EB, and AH to
HD, HE is parallela to DB: For the

a 2. 6. same reason, HK is parallel to AB: Therefore HEBK is a parallelogram,

E and HK equal to EB: but EB is equal

b 34. 1. to AE; therefore also AE is equal to

B F HK: And AH is equal to HD; wherefore EA, AH are equal to KH, HD, each to each; and the angle EAH is equal to the angle c 29. 1. KHD; therefore the base EH is equal to the base KD, and

. Because since a fourth proportional to the squares of BD, FH, and the circle ABCD, is possible, and since it can neither be less nor greater than the circle EFGH, it inust be equal to it.

« ForrigeFortsett »