Sidebilder
PDF
ePub

IF

PROP. XL. THEOR.

there be two triangular prifms of the fame altitude, the base of one of which is a parallelogram, and the base of the other a triangle; if the parallelogram be double of the triangle, the prifms fhall be equal to one another.

Let the prifms ABCDEF, GHKLMN be of the fame altitude, the firft whereof 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 one of them have a parallelogram AF, and the other a triangle GHK for its base; if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prifm GHKLMN.

Book XI.

Complete the folids AX, GO; and because the parallelogram AF is double of the triangle GHK; and the parallelogram HK double 2 of the fame triangle; therefore the parallelogram AF is a. 34. 1.

[blocks in formation]

equal to HK. but folid parallelepipeds upon equal bases, and of

the fame altitude are equal ↳ to one another. therefore the folid b. 31. 11. AX is equal to the folid GO. and the prism ABCDEF is half c c. 28. 11. of the folid AX; and the prism GHKLMN half of the folid GO.

therefore the prism ABCDEF is equal to the prifm GHKLMN. Wherefore if there be two, &c. Q. E. D.

[blocks in formation]

Which is the firft Propofition of the tenth Book, and is neceffary to fome of the Propofitions of this Book.

[ocr errors]

F 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 less than the leaft of the proposed magnitudes.

A

D

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 fo on; there fhall at length remain a magnitude lefs than C.

H

F

For C may be multiplied fo as at length to K become greater than AB. let it be fo multiplied, and let DE its multiple be 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 fo on until there be as many divifions in AB as there are in DE. and let the divifions AK, KH, HB be as many as the divifions DF, FG, GE. and because DE is greater than AB, and that EG taken from DE is lefs than its

В CE

half, but BH taken from AB is greater than its half, therefore the Book XII. remainder GD is greater than the remainder HA. again, because GD is greater than HA, and that GF is 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 fame thing may in the fame way be demonftrated.

SIM

PROP. I. THEOR.

IMILAR 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.

a

Join BE, AM, GL, FN. and because the polygon ABCDE is fimilar to the polygon FGHKL, the angle BAE is equal a to the angle a. 1. Def. 6. GFL, and as BA to AE, fo is GF to FL. wherefore the two triangles BAE, GFL having one angle in one equal to one angle in the

[blocks in formation]

other, and the fides about the equal angles proportionals, are equi

c

angular; and therefore the angle AEB is equal to the angle FLG. b. 6. 6. but AEB is equal to the angle AMB, because they stand upon the c. 21. 3a fame circumference; and the angle FLG is, for the fame 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 GFN; d. 31. 3. wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another. therefore as BM to

Book XII. GN, fo 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

e. 4. 6.

[ocr errors]

10. Def.5.

and 22.5.

g. 20. 6.

8 ratio of that which BM has to GN; and the ratio of the polygon

[blocks in formation]

See N.

3. 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 fimilar polygons, &c. Q. E. D.

C

PROP. II. THEOR.

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

Let ABCD, EFGH be two circles, and BD, FH their diame ters. 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 square of FH, as the circle ABCD is to fome space 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 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 to the circle, the square EFGH is half a of the fquare described

*For there is fome fquare equal to the circle ABCD; let P be the fide of it. and to three straight lines BD, FH and P, there can be a fourth proportional, let this be Q. therefore the fquares of thefe four ftraight lines are propor

tionals; that is, to the fquares 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.

about the circle; and the circle is lefs than the fquare described 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 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 half of the parallelogram in which it is. a. 41. 1. but every segment 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

[blocks in formation]

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 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 lcaft 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 excess of the circle EFGH above S. therefore the rest 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 fquare of BD is to the square of FH, fobis the polygon AXBOCPDR to the polygon LKFLGMHN, b. 1. 12. but the fquare of BD is alfo to the fquare of FH, as the circle ABCD

« ForrigeFortsett »