e

Book XII. GN, fo is BA to GF, and therefore the duplicate ratio of BM to GN, is the fame with the duplicate ratio of BA to GF. but the ratio of the fquare of BM to the fquare of GN, is the duplicate 8 ratio of that which BM has to GN; and the ratio of the polygon

€ 4. 6

10. Def S.

and 22.5.

See N.

2. 41. I.

ABCDE to the polygon FGHKL is the duplicate of that which
BA has to GF. therefore as the fquare of BM to the fquare
of GN, fo is the polygon ABCDE to the polygon FGHKL.
Wherefore fi.nilar polygons, &c. Q. E. D.

C

PROP. II. THEO R.

IRCLES are to one another as the fquares of their diameters.

Let ABCD, EFGH be two circles, and FD, FH their diameters. as the fquare of B1) to the fquare of 1H, to 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 ALCD is to fome pace either less than the circle EFGH, or greater than it *. Fift, let it be to a space S lefs-than the circle EI GH; and in the circle EFGH defcribe the fquare HIGH. this fquare is greater than half of the circle EFGH; becaufe if through the points E, F, G, H, there be drawn tangents to the circle, the fquare LFGH is half of the fquare defcribed

For there is fome fquare equal to the circle ABCD; let P be the fice of it. and to three ftraight lines ED, FH and P, there can be a fourth proportional, let this be Q. therefore the squares of there tour straight lines are proportio

nals; that is, to the fquares of ED, FH and the circle ABCD it is poff le there may be a fourth proportional. Let this be S. and in like manner are o be understood fome things in fome of the following Propofitions.

about the circle; and the circle is less than the fquare defcribed about Book XII. it; therefore the fquare EFGH is greater than half of the circle. Divide 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 straight 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 shall be the half of the parallelogram in which it is. a. 41. 1. but every fegment is less 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

a

this, there will at length remain fegments of the circle which toge ther fhall be lefs than the excefs of the circle EFGH above the fpace 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 fo on, there fhall at length remain a magnitude lefs than the least of the propofed magnitudes. Let then the fegments EK, KF, FL, LG, GM, MH, HN, NE be thofe that remain and are together lefs than the excefs of the circle EFGH above S. therefore the reft of the circle, viz. the polygon EKFLGMHN is greater than the space S. Defcribe likewife in the circle ABCD the polygon AXBOCPDR fimilar to the polygon EKFLGMHN. as, therefore, the fquare of BD is to the fquare of FH, fob is the polygon AXBOCPDR to the polygon EKFLGMHN. b. 1. 13, but the fquare of BD is alfo to the fquare of FH, as the circle ABCD

Book XII.

C. 11. 5.

d. 4. 5.

C

is to the space S. therefore 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 the space S is greater than the polygon EKFLGMHN. but it is likewife lefs, as has been demonftrated; which is impoffible. Therefore the fquare of BD is not to the fquare of FH, as the circle ABCD is to any space lefs 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 fpace 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 so to T a space greater than the circle EFGH. therefore, inverfely, as the fquare of FH to the fquare of BD, so is

the fpace T to the circle ABCD. but as the fpace † T is to the circle ABCD, fo is the circle EFGH to fome fpace, which muft be lefs than the circle ABCD, because the space T is greater, by hypothefis, than the circle EFGH. therefore as the fquare of FH is to

For as in the foregoing Note at
it was explained how it was poffible
there could be a fourth proportional
to the fquares of BD, FH and the circle
ABCD, which was named S. fo 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 fome of the following Propofitions.

the fquare of BD, fo is the circle EFGH to a space less than the Book XII. circle ABCD, which has been demonftrated to be impoffible. therefore the fquare of BD is not to the fquare of FH, as the circle ABCD is to any space 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 fpace lefs than the circle EFGH. wherefore as the fquare of BD to the fquare of FH, fo is the circle ABCD to the circle EFGH+. Circles, therefore, are, &c. Q. E. D.

EVERY pyramid having a triangular bafe, may be See N.

divided into two equal and fimilar pyramids having triangular bafes, and which are fimilar to the whole pyramid; and into two equal prifms 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 haying 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. Because AE is equal to EB, and AH to HD, HE is parallel 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 B equal to HK. and AH is equal to HD;

K

E

F

L

T

C

wherefore EA, AH are equal to KH, HD, each to each; and the

c

2. 2.

b. 34. I

angle EAH is equal to the angle KHD; therefore the bafe EH is c. 29. 1.

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

d. 4. I.

e. 10. 11.

f. C. 11.

e

D

Book XII. equal to the bafe KD, and the triangle AEH equal and fimilar to the triangle HKD. for the fame reason, the triangle AGH is equal and similar to the triangle HLD. and 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 fame plane with them, they contain equal angles; therefore the angle EHG is equal to the angle KDL. again, because EH, HG are equal to KD, DL, each to each, and the angle EHG equal to the angle KDL; therefore the bafe EG is equal to the base KL, and the triangle EHG equal and similar to the triangle KDL. for the fame reafon, the triangle AEG is alfo equal and fimilar to the triangle HKL. Therefore the pyramid of which the bafe is the triangle AEG, and of which the vertex is the point H, is equal f and fimilar to the pyramid the bafe of which is the triangle KHL, and vertex the point D. and becaufe HK is parallel to AB a fide of the triangle ADB, the triangle ADB is equiangular to the triangle HDK, and their fides are proportionals . therefore the triangle ADB is fimilar to the triangle HDK. and for the fame reafon, the triangle DBC is fimilar to the triangle DKL; and the triangle ADC to the triangle HDL; and alfo the triangle ABC to the triangle AEG. but the triangle AEG is fimilar to the triangle HKL, as before was proved,

8.4 6.

h. 21. 6.

B

K

E

F

C

therefore the triangle ABC is fimilar to the triangle HKL. and the pyramid of which the bafe is the triangle ABC, and vertex the point i. B. 11. & D, is therefore fimilar i to the pyramid of which the bafe is the tri11.Def.. angle IIKL, and vertex the fame point D. but the pyramid of which

k. 41. I.

the bafe is the triangle HKL, and vertex the point D,is fimilar, 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 fimilar to the pyramid of which the base is the triangle AEG and vertex H. therefore each of the pyramids AEGH, HKLD is fimilar to the whole pyramid ABCD. and becaufe BF is equal to FC, the parallelogram EBFG is double k of the triangle GFC. but when there are two prifms of the fame altitude, of which one has a parallelogram for