that EG taken from DE is not greater than its half, but BH ta- Book XII. ken 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 that 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 lefs than C. Q. E. D.

And if only the halves be taken away, the fame thing may in the fame way be demonftrated.

PROP. I. THEOR.

SIMILAR polygons infcribed in circles, are to one another as the fquares of their diameters.

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

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

c

quiangular b; and therefore the angle AEB is equal to the angle b 6. 7. FLG: But AEB is equal to the AMB, because they ftand up- c 21. 3. on the fame circumference; and the angle FLG is, for the fame reafon, equal to the angle FNG: Therefore alfo the angle AMB is equal to FNG: And the right angle BAM is equal to the right a angle GFN; wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one

another :

d3r. 3

€ 4. 6.

Book XII. another: Therefore as BM to GN, focis BA to GF and there 'fore the duplicate ratio of BM to GN, is the fame f with the du f1o. def. plicate ratio of BA to GF: But the ratio of the fquare of BM to s. & 22. the fquare of GN, is the duplicate g ratio of that which BAI has to GN; and the ratio of the polygon ABCDE to the polygon

5. g. 20. 6.

"

See N.

FGHKL is the duplicate 8 of that which BA has to GF: There: fore as the fquare of BM to the fquare of GN, fo is the polygon ABCDE to the polygon FGHKL. Wherefore fimilar polygons, &c. Q. E. D.

a

CIRCLES are to one another as the fquares of their

diameters.

Let ABCD, EFGH be two circles, and BD, FH their dia meters: As the fquare of BD to the fquare of FH, fo is the circle ABCD to the circle EFGH.

For, if it be not fo, the fquare of BD fhall be to the fquare of FH, as the circle ABCD is to fome fpace either less than the circle EFGH, or greater than it *. First, let it be to a space S less than the circle EFGH; and in the circle EFGH defcribe the fquare EFGH: This fquare is greater than half of the circle EFGH; because if, through the points E, F, G, H, there be drawn tangents to the circle abe fquare

*For there is some fquare equal to the circle ABCD; let P be the fide of it, and to three ftraight lines BD, FH and P, there can be a fourth proportional; let this be Q: Therefore the fquares of these four straight lines are

proportionals; that is, to the squares of BD, FH and the circle ABCD, it is poffible there may be a fourth proportion al. Let this be S. And in like man ner are to be understood fome things in fome of the following propofitions.

fquare EFGH is half of the fquare defcribed about the circle; Book XII. and the circle is lefs than the fquare defcribed about it; therefore the fquare EFGH is greater than half of the circle. Di-2 47. z. vide 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 of the fegment of the circle it ftands in; because if ftraight lines touching the circle be drawn through the points K, L, M, N, and parallelograms upon the ftraight lines EF, FG, GH, HE be completed; each of the triangles EKF, FLG, GMH, HNE fhall be the half aa 41. I. of the parallelogram in which it is: But every fegment is lefs than the parallelogram in which it is: Wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the fegment of the circle which contains it: And if thefe circumferences before named be divided each into two equal parts, and their extremities be joined by ftraight lines, by continuing

to do this, there will at length remain fegments of the circle which, together, fhall be lefs than the excefs of the circle EFGH above the space S: Becaufe, 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 fhall at length remain a magnitude lets than the leaft of the propofed magnitudes. Let then the feg. ments EK, KF, FL, LG, GM, MH, HN, NE be thofe that remain and are together less than the excess of the circle EFGH above S: Therefore the reft of the circle, viz. the polygon EKFLGMHN, is greater than the fpace S. Defcribe likewife in the circle ABCD the polygon AXBOCPDR fimilar to the polygon EKFLGMHN: As, therefore, the quare of BI) is to the fquare of FH, fo b is the polygon AXBOCPDR to the po-b 1. 12. lygon EKFLGMHN: But the fquare of BD is alfo to the

fquare

Book XII. fquare of FH, as the circle ABCD is to the space S: There. 'fore as the circle ABCD is to the fpace S, fo is the polygon AXBOCPDR to the polygon EKFLGMHN: But the circle ABCD is greater than the polygon contained in it; wherefore 145 the space S is greater d than the polygon EKFLGMHN: But

it is likewife lefs, as has been demonftrated; which is impoffi ble. Therefore the fquare of BD is not to the fquare of FH, as the circle ABCD is to any space less than the circle EFGH. In the fame manner, it may be demonftrated, that neither is the fquare of FH to the fquare of BD, as the circle EFGH is to any space less than the circle ABCD. Nor is the fquare of BD to the fquare of FH, as the circle ABCD is to any space greater than the circle EFGH: For, if poffible, let it be fo to T, a space greater than the circle EFGH: Therefore, inversely, as the fquare of FH to the fquare of BD, fo is the space T to

the circle ABCD. But as the space + T ́is to the circle ABCD, fo is the circle EFGH to fome fpace, which must be lefs d than the circle ABCD, because the space T is greater, by hypothe fis, than the circle EFGH. Therefore as the fquare of FH is to the

the fquare of BD, fo is the circle EFGH to a fpace lefs than Book XII. the circle ABCD, which has been demonstrated to be impoffible: Therefore the fquare of BD is not to the fquare of FH, as the circle ABCD is to any fpace greater than the circle EFGH: And it has been demonftrated, that neither is the fquare of BD to the fquare of FH, as the circle ABCD to any space lefs than the circle EFGH: Wherefore, as the fquare of BD to the fquare of FH, to 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 di- See N. vided into two equal and fimilar pyramids having triangular bafes, and which are fimilar to the whole pyramid; and into two equal prifins which together are greater than half of the whole pyramid.

D

Let there be a pyramid of which the bafe is the triangle ABC and its vertex the point D: The pyramid ABCD may be divided into two equal and fimilar pyramids having triangular bafes, and fimilar to the whole; and into two equal prifms which together are greater than half of the whole pyramid.

Divide AB, BC, CA, AD, DB, DC, each into two equal parts in the points E, F, G, H, K, L, and join EH, EG, GH, HK, KL, LH, EK, KF, FG. Becaule AE is equal to EB, and AH to HD, HE is parallel a to DB: For the fame reafon, HK is parallel to AB: Therefore HEBK is a parallelogram, and HK equal to EB: But EB is equal to AE; therefore alfo AE is equal to HK: And AH is equal to HD; wherefore EA, AH are equal to KH, HD,

K

B

E

a 2.

G

b 34. I.

F

each to each; and the angle EAH is equal e to the angle KHD ; ́c 29. 1. therefore the bafe EH is equal to the bafe KD, and the triangle

Ꭱ

AEH

Because as a fourth proportional to the fquares of BD, FH and the circle ABCD is poffible, and that it can neither be lefs nor greater than the circle ESGH, it must be equal to it.