D

K

F

Y

E

L

T

M

H

XY to YO, and (I. 29.) the contained angles DXY, YOE are equal; DY is equal (I. 4.) to YE, and the angle XYD to OYE; and therefore DYE (1. 14. cor.) is a straight line. For the same reason, BSG is a straight line, and BS equal to SG. Again, because CA is equal and parallel to DB, and also to EG; therefore (I. ax. 1. and XI. 9.) DB is equal and parallel to EG; and DE, BG join their extremities: therefore (I. 33.) DE is equal and parallel to BG: also DG, YS are drawn from points in the one, to points in the other; they are therefore in one plane, and must meet one another: let them meet in T. Then, because DE is parallel to BG, the alternate angles EDT, BGT are (I. 29.) equal; also (I. 15.) the angles DTY, GTS are equal; and therefore in the triangles DTY, GTS, since the sides DY, GS are equal, being halves of DE, BG; DT is equal (I. 26.) to TG, and YT equal to TS. Wherefore, in a parallelepiped, &c.

B

P

PROP. XL. THEOR.

A

N

R

Ir there be two triangular prisms of the same altitude, the base of one of which is a parallelogram, and that of the other a triangle; if the parallelogram be double of the triangle, the prisms are equal.

Let the prisms ABCDEF, GHKLMN be of the same altitude, the first of which is contained by the two triangles ABE, CDF, and the three parallelograms AD, DE, EC; and the other by the two triangles GHK, LMN and the three parallelograms LH, HN, NG; and let the parallelogram AF be taken as the base of one of them, and the triangle GHK, as base of the other; if AF be double of GHK, the A prisms are equal.

E

R

D

X

H

M

N

Complete the parallelepipeds AX, GO; and because the parallelogram AF is double of the triangle GHK; and (I. 34.) the parallelogram HK double of the same triangle; therefore the bases AF, HK are equal. But (XI. 31.) parallelepipeds upon equal bases, and of the same altitude, are equal to one another. Therefore AX is equal to GO; and (XI. 28.) the prism ABCDEF is half of AX; and the other prism half of GO. Therefore (I. ax. 7.) the prisms are equal. Wherefore, if there be two triangular prisms, &c.

BOOK XII.

LEMMA I.*

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 will at length remain a magnitude less than the least of the proposed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more than its half, and from the remainder more than its half, and so on; there will at length remain a magnitude less than C.

K

A

For C may be multiplied so as at length to become greater than AB. Let it be so multiplied, till DE, its multiple, is greater than AB, and let DE be divided into DF, FG, GE, each equal to C. From AB take BH greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: and H let the divisions in AB be AK, KH, HB; and the divisions in ED be DF, FG, GE. Then, because DE is greater than AB, and that EG taken from DE C E is not greater than its half; but BH 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 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 AK is less than C.

B

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

This, which is the first proposition of the tenth book, is necessary for some of the propositions of this book.

PROP. I. THEOR.

SIMILAR polygons inscribed in circles, are to one another as the squares of the diameters.

In the circles ACE, FHL, let there be the similar polygons ABCDE, FGHKL; and let BM, GN be the diameters of the circles the square of BM is to the square of GN as the polygon AC is to the polygon FH.

:

B

E

G

L

M

N

H

K

Join BE, AM, GL, FN. Then, because the polygons are similar, and (VI. 20.) similar polygons are divided into similar triangles; the triangles ABE, FGL are similar, and (VI. def. 1.) equiangular; the angle AEB therefore is equal to FLG. But (III. 21.) AEB is equal to AMB, because they stand on the same arc; and for the same reason, the angles FLG, FNG are equal: therefore also the angles AMB, FNG are equal: and the right angles (III. 31.) BAM, GFN are equal; therefore (VI. 4. schol. and alternately) BM: GN:: BA: GF; and therefore (ex æquo, and V. def. 11.) the duplicate ratio of BM to GN, is the same with the duplicate ratio of BA to GF. But (VI. 20.) the ratio of the square of BM to the square of GN, is the duplicate ratio of that which BM has to GN; and the ratio of the polygon AC to the polygon FH, is the duplicate of that which BA has to GF. Therefore (V. 11.) the square of BM is to the square of GN, as the polygon AC to the polygon FH. Wherefore similar polygons, &c.

PROP. II. THEOR.

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

Let AC, EG be two circles, and BD, FH their diameters: as the square of BD to the square of FH, so is AC to EG. For, if it be not so, the square of BD will be to the square of FH, as the circle AC is to some space either less than the circle EG, or greater than it. First, let it be to a space S less than EG; and in EG describe the square EFGH. This square is greater than half the circle EG; because, if through the points E, F, G, H, there be drawn tangents to the circle, the square EFGH is half (I. 41.) of the square described about the circle; and the circle is less than the square described about it; therefore the square EFGH is greater than half the circle. Bisect the arcs EF, FG, GH, HE, in the points K, L, M, N, and join EK, KF, &c. Then each of the triangles EKF, FLG, &c., is greater than half of the segment of the circle in which it stands; because, if straight lines touching the circle be drawn through the points K,

L, M, N, and parallelograms upon the straight lines EF, FG, &c., be completed, each of the triangles EKF, FLG, &c., is half (I. 41.) 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, &c., is greater than half the segment of the circle which contains it. And if each of the arcs EK, KF, &c., be bisected, and their extremities be joined by straight lines, by continuing to do this, there will at length remain segments of the circle which, by the preceding lemma, will be together less than the excess of the circle EFGH above the space S. 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: then the rest of the circle, viz., the poly

gon EKFLGMHN, is greater than S. Describe likewise in the circle AC the polygon AXBOCPDR similar to the polygon EKFLGMHN: then (XII. 1.) as the square of BD is to the square of FH, so is the first of these polygons to the second. But the square of BD is also to the square of FH, as the circle AC is to S: therefore (V. 11.) as the circle AC is to S, so is the polygon AXBOCPDR to the polygon EKFLGMHN. But the circle AC is greater than the polygon contained in it; wherefore (V. 14.) the space S is greater 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 AC is to any space less than the circle EG. In the same manner it may be demonstrated, that neither is the square of FH to the square of BD, as the circle EG is to any space less than the circle AC. Nor is the square of BD to the square of FH, as the circle AC is to any space greater than the circle EG. For, if possible, let it be so to T, a space greater than the circle EG: therefore inversely as the square of FH to the square of BD, so is the space T to the circle AC. But as T is to the circle AC, so is the circle EG to some space, which (V. 14.) must be less than the circle AC, because T is greater, by hypothesis, than the circle EG. Therefore as the square of FH is to the square of BD, so is the circle EG to a space less than the circle AC, which has been demonstrated to be impossible. Therefore the square of BD is not to the square of FH, as the circle AC is to any space greater than the circle EG: and it has been demonstrated, that neither is the square of BD to the square of FH, as the circle AC to any space less than the

P

circle EG: wherefore, as the square of BD to the square of FH, so is the circle AC to the circle EG. Circles, therefore, &c.

PROP. III. THEOR.

A TRIANGULAR pyramid may be divided into two equal and similar triangular pyramids, 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 the base is the triangle ABC, and the vertex D: the pyramid ABCD may be divided into two equal and similar triangular pyramids, which are similar to the whole; and into two equal prisms which together are greater than half of the whole pyramid.

B

H

K

F

G

Bisect AB, BC, CA, AD, DB, DC, 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 (VI. 2.) to DB. For the same reason, HK is parallel to AB; and therefore HEBK is a parallelogram, and (I. 34.) HK is equal to EB. But EB is equal to AE; therefore also AE is equal to HK: and AH is equal to HD; also (I. 29.) the angle EAH is equal to KHD; therefore (I. 4.) in the triangles HAE, DHK, EH is equal to KD, and the triangle AEH is equal and similar to the triangle HKD. For the same reason, the triangle AGH is equal and similar to HLD: and (XI. 10.) because the two straight lines EH, HG are parallel to KD, DL, the angles EHG, KDL are equal: also EH, HG are equal to KD, DL, each to each; and therefore (I. 4.) EG is equal to KL, and the triangle EHG equal and similar to KDL. For the same reason, the triangle AEG is also equal and similar to HKL. Therefore (XI. C.) the pyramid AEGH, of which the base is the triangle AEG, and the vertex H, is equal and similar to the pyramid HKLD, the base of which is the triangle KHL, and the vertex D. Also, because HK is parallel to AB, a side of the triangle ADB, the triangle ADB is equiangular to the triangle HDK, and (VI. 4.) their sides are proportionals: they are therefore similar. For the same reason, the triangles DBC, DHL are similar: so also are the triangles ADC, HDL; and ABC, AEG. But the triangles AEG, HKL, have been proved to be similar: therefore (VI.21.) the triangle ABC is similar to HKL. The pyramid ABCD is therefore similar (XI. B. and def. 8.) to the pyramid HKLD. But the pyramid HKLD is similar, as has been proved, to the pyramid AEGH; wherefore the pyramid ABCD, is similar to the pyramid AEGH. Therefore each of the pyramids