Book XII. GN, foc is BA to GF, and therefore the duplicate ratio of BM to GN, is the fame f with the duplicate ratio of BA to GF. but the ratio of the fquare of BM to the fquare of GN, is the duplicate & ratio f. 10.Def.5 of that which BM has to GN; and the ratio of the polygon ABCDE

c. 4. 6.

and 22. 5. g. 20. 6.

Sce N.

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, so is the polygon ABCDE to the polygon FGHKL. Wherefore fimilar polygons, &c. Q. E. D.

PROP. II. THEOR.

CIRCLES
Es are to one another as the fquares of their dia-

meters.

Let ABCD, EFGH be two circles, and BD, FH their diameters. 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 space either lefs than the circle EFGH, or greater than it *. First, let it be to a fspace S less than the circle EFGH; and in the circle EFGH describe the square EFGH. this fquare is greater than half of the circle EFGH; becaufe if through the points E, F, G, H, there be drawn tangents a. 41. 1. to the circle, the fquare EFGH is half of the square described a

For there is fome fquare equal to the circle ABCD; let P be the side 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 thefe four ftraight lines are proportio

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

bout

bout the circle; and the circle is less than the fquare defcribed aboutBook XH. it; therefore the square 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 straight 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 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 these circumferences before named be divided each into two equal parts, and their extremities be joined by straight lines, by continuing to do

A

a

X

B

C

this, there will at length remain fegments of the circle which together fhall be lefs than the excefs of the circle EFGH above the fpace S. because, by the preceeding 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 leaft of the propofed magnitudes. Let then the fegments EK, KF, FL, LG, GM, MH, HN, NE be those that remain and are together lefs than the excefs of the circle EFGH above S. therefore the rest 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 square of BD is to the fquare of FH, fob is the polygon AXBOCPDR to the polygon EKFLGMIN, b. 1. 12. but the fquare of BD is alfo to the fquare of FH, as the circle ABCD

Book XII.is to the space S. therefore as the circle ABCD is to the space S, fo is the polygon AXBOCPDR to the polygon EKFLGMHN. but c. 11. 5. the circle ABCD is greater than the polygon contained in it; whered. 14. 5. fore the fpace 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 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 fpace greater than the circle EFGH. therefore, inversely, 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 must be lefs than the circle ABCD, becaufe the fpace 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 fpace, named T, and the circles ABCD, EFGH, and the like is to be understood in fome of the following Propofitions.

the

the fquare of BD, fo is the circle EFGH to a fpace lefs 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 fspace 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 less 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.

EVE

PROP. III. THEOR.

VERY 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.

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. Because AE is equal to EB, and AH to HD, HE is parallel to DB. for the fame reason, HK is parallel to AB. therefore HEBK is a parallelogram, and HK equal b to EB. but EB is equal to AE; therefore alfo AE is B equal to HK. and AH is equal to HD;

K

Ꭰ .

H

L

E

G

b. 34. I.

F

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

C

angle EAH is equal to the angle KHD; therefore the base 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.

equal

Book XII. equal to the base KD, and the triangle AEH equal and fimilar to the

triangle HKD. for the fame reason, the triangle AGH is equal and fimilar to the triangle HLD. and because the two straight lines EH,

HG which meet one another are parallel to KD, DL that meet one e. 10. 11. another, and are not in the fame plane with them, they contain equale 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 base EG is equal to the base KL, and the triangle EHG equal and fimilar to the triangle KD L. 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

f. C. 11. point H, is equal f and fimilar to the py

8.4. 6.

ramid the bafe of which is the triangle.
KHL, and vertex the point D. and be-
caufe HK is parallel to AB a fide of the
triangle ADB, the triangle ADB is equi-
angular to the triangle HDK, and their
fides are proportionals . therefore the
triangle ADB is fimilar to the triangle K
HDK. and for the fame reason, the tri-
angle 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 B
was proved, therefore the triangle ABC is

E

h. 21. 6. fimilar to the triangle HKL. and the pyramid of which the base is i. B. 11. & the triangle ABC, and vertex the point D, is therefore fimilar i to 11.Def.11. the pyramid of which the bafe is the triangle HKL, and vertex the

fame point D. but the pyramid of which the base 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 bafe is the triangle AEG and vertex H. therefore each of the pyramids AEGH, HKLD is fimilar to the whole pyramid ABCD. and bek. 41. 1. caufe 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 its bafe, and the other

a tri