Let the plane

be together greater than the third ABC be perpendi

CAD. cular to DBC, and

If one of the angles B AC, BAD let any straight line

be greater than CAD, or equal to it, AB be drawn in

the proposition is evident; for this, tothe plane ABC per

gether with the other, must be greater pendicular to the

than AD. section

But, if the angles BAC, BAD be each BC: the straight

of them less than C A D, from the angle line A B shall be perpendicular to the CAD cut off the angle CAE equal to plane D BC. From the point B, in BAC; in A B take any point B, make the plane D B C, let B D be drawn at A E equal to A B; through E draw right angles to BC. Then, because the any line CED, not parallel to A C or planes are at right angles to one ano- AD, and therefore cutting A C, A D in ther, the angle ABD is a right angle the points C, D respectively; and join (17. Cor.): but A B C is likewise a BC, BD. Then, because the triangles right angle; therefore (3.) A B is per- A B C, AEC have two sides of the pendicular to the plane D BC. one equal to two sides of the other,

Next, let the straight line A B be per- each to each, and the included angles pendicular to the plane B CD; and let B AC, E A C equal to one another, A B C be any plane passing through the base B C is equal to the base EC AB: the plane A B C shall be perpen- (I. 4.): but B D and B C are together dicular to BCD.

greater than DC, (I. 10.) that is, than Let BC be the common section of the D E and EC: therefore B D is greater two planes; and, from the point B, in than E D. And, because in the trianthe plane BCD, draw BỐ at right gles ABD, A ED the two sides A B, angles to BC. Then, because AB AD of the one are equal to the two sides meets the line BD drawn in the plane AE, AD of the other, each to each, but to which A B is perpendicular, the angle the base B D greater than the base E D, ABD is a right angle (def. 1.). But the angle BAD is greater than the the angle ABD is contained by straight angle E AD(I. 11.). Therefore, B A C, lines drawn in the two planes perpen- BAD together are greater than EAC, dicular to their common section B C. EAD together, that is, than CAD. Therefore (17. Cor.) the dihedral angle Therefore, &c. A B C D is likewise a right dihedral Cor. If three plane angles form a angle, and the plane A B C is perpen- solid angle, any one of them shall be dicular to B C D.

greater than the difference of the other Therefore, &c.

two. Cor. 1. If through the same point there pass any number of planes per- PROP. 20. (Euc. xi. 21.) pendicular to the same plane, they shall all of them pass through the same

If a solid angle be contained by any straight line, viz. the perpendicular number of plane angles; these shall be which is drawn from the point to the together less than four right angles. plane.

Let the solid anCor. 2. Euc. xi. 19. If two planes, gle at A be conwhich cut one another, be each of them tained by any numperpendicular to a third plane, their ber of plane angles common section shall be perpendicular BAC, CAD,DĂE, to the same plane.

EAF, FAB: these

shall be together Prop. 19. (Euc. xi. 20.)

less than four right If a solid angle be contained by three angles. plane angles, any two of these shall be Let the edges of the solid angle be cut, together greater than the third.

by any the same plane, in the points B,C, Let the solid an

D, E, F respectively, and therefore the gle at A be con

planes which contain it in the straight tained by the three

lines B C, CD, DE, E F, FB (2.), which plane angles BAC,

together contain the rectilineal figure BAD, CAD: any

BCDEF. Then, because the triangles two of them, as

ABC, AC D, &c. are as many, in numBAC, BAD, shall

ber, as the sides of the figure, the angles



of these triangles are (I. 19.) together that they are contained by 4, 8, 20, 6, equal to twice as many right angles as and 12 faces respectively. the figure has sides. But the angles of The further consideration of solid these triangles are those which contain angles is deferred to Book vi., where we the solid angle at A, together with the shall find that the investigation of their pairs of angles at the points B, C, D, properties is greatly facilitated by means &c.; and twice as many right angles as of the spherical surface described about the figure has sides are equal to four the angular point, and that they are, right angles together with the angles indeed, perfectly analogous to the pro. of the figure (I. 20.). Therefore, the perties of triangles upon such a surface. angles at A, together with the pairs of angles at B, C, D, &c. are equal to four

SECTION 3.-Of Solids contained by right angles together with the angles of

Planes. the figure BCDEF. But, because the solid angles at B, C, D, &c. are each

PROP. 21. of them contained by three plane angles, If two triangular prisms have a prinany two of which are (19.) together cipal edge of the one equal to a principal greater than the third, the pairs of an edge of the other, and the dihedral angles at B, C, D, &c. are together greater gles at those edges equal to one another, than the angles of the figure B C D E F. and likewise the sides which contain Therefore, the angles at A are together those dihedral angles equal, each to less than four right angles.

each ; the two prisms shall be equal to Therefore, &c.

one another. Cor. It follows from this proposition

Let A B C D E F, ab c d e f be two trithat there cannot be more than five regu. angular prisms which have the

edge AB lar solids. For the solid angles of a regu- equal to the edge ab, the dihedral angle lar solid are contained by the plane angles of equilateral triangles, or of squares,

DABE equal to the dihedral angle or of regular pentagons, &c. (def

. 9.). the sides ac, af each to each: the prism

dabe, and the sides A C, A F equal to Now, six angles of equilateral triangles ABCDEF shall be equal to the prism are together equal to four right angles

a b c d ef. (I. 19.): therefore, only three, or four, or five, of such angles, may be taken to form a solid angle; and, accordingly, there cannot be more than three different regular solids, whose faces are equilateral triangles. Again, three angles of squares may be taken to form a solid angle ; but four angles of squares are equal to four right angles First, let the angle A B C be equal to (I. def. 20.). * Lastly, three angles of the angle abc, and the angle AB F to regular pentagons may be taken to form the angle abf, so that the sides AF, a solid angle, for these are together less AC are similar and similarly placed (I. 20.) than four right angles. But, be to the sides af, ac, each to each. Then, cause the angle of a regular pentagon is if the straight line A B be made to coin: greater (I. 20.) than that of a square, cide with the straight line ab, which is four angles of a regular pentagon are equal to it, and the plane AF with the greater than four right angles, so that plane af, the plane AC will also coinnot more than three of them can be taken cide with the plane ac, because the to form a solid angle. With regard to dihedral angle D ABE is equal to the the regular hexagon, and other regular dihedral angle dabe: also the lines figures that have a still greater number BC, B F will coincide with the lines b c, of sides, three angles of a regular hexa, bf respectively; and therefore the lines gon are equal to four right angles, and AD, A E, which are parallel and equal those of the others still greater (I. 20.). to the former, each to each, will coinIt appears, therefore, that there cannot cide respectively with the lines ad, ae, be more than five regular solids; of which are parallel and equal to the which, if there be so many, three will be latter: therefore the prism ABCDEF included by equilateral triangles, one by will coincide with the prism a b cde f. squares, and one by regular pentagons. And because the two prisms may be It will be seen in the section of problems, made to coincide, they are equal to one that all of these may be constructed, and another (I. ax, 11.).


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Next, let the sides A C, ac be equal and complete the triangular prism AB ·and similar, and the sides A F, af equal CDH G, as in the preceding case. Then, but dissimilar. Make the angle A BG, it is evident that the prism ABCDHG in the plane A BF, equal to the angle is equal to each of the prisms ABCD abf; complete the parallelogram A B E F, abcdef; to the former, because GH; and join CG, D H. Then because the side AC is common to both, and the GH is equal and parallel to AB (I. 22.), other sides AG, AF are equal to one anowhich is equal and parallel to CD, ther; and to the latter, because the sides the three, AB, CD, GH are equal AG, af are equal and similar (I. 24. and parallel, and, therefore, (def. 14.) and I. ax. 1.), and the other sides AC, ac ABCDHG is a triangular prism. equal to one another. Therefore (I. ax. 1.)

the prism ABCDEF is equal to the
prism abcdef.
Therefore, &c.

PROP, 22. (Euc. xi. 24 and 28.)
The opposite faces of a parallelopiped
are similar and equal parallelograms :

also any two of its opposite edges are Also, the parallelogram AHGB is equal parallel to one another; and the plane to the parallelogram A EFB (I. 24.), which passes through them bisects the that is, to the parallelogram a efb; parallelopiped. and the parallelogram ADCB is equal Let A a be a paralto the parallelogram ad cb, by the lelopiped, and CD, supposition. Therefore, because the cd any two opposite sides AC, A G of this new prism are faces; they shall be respectively equal, similar, and similarly similar and equal paplaced to the sides ac, a f of the prism rallelograms. They a b c d e f, it is equal to the prism are parallelograms, for the opposite sides abcdef, by the former case. Again, of each, as A C, Db, being sections of because H G is equal to D C, that is to parallel planes by the same plane, are EF, HE is equal to GF, and may be parallel (12). Again, their made to coincide with it; also the di- sponding sides, as AD, B c, are equal hedral angle A EHD is equal to the to one another, because they are oppodihedral angle BFGC, and may be site sides of a parallelogram ABCD made to coincide with it; in which case, (I. 22.); and their corresponding angles, also, the angles HED, HE A will co- as CAD, dBc are equal to one another, incide with the angles GFC, GFB, because they are intercepted upon pawhich (I. 15.) are equal to them respec- rallel planes C AD, d B C by the planes tively, and the lines E D, E A with the of the same dihedral angle CA BD lines FC, FB, which are equal to them (15. Cor.): therefore the parallelograms respectively (I. 22.). Therefore the CD, cd are equal and similar. solid HEAD, which is bounded by the

Also, any two of the opposite edges of four triangular planes H D E, HD A, the parallelopiped, A C, ac, are paHEA, E A D, may be made to coincide rallel to one another; for they are the with the solid GFBC, which is bounded

common sections of parallel planes, by the four triangular planes GCF, namely, the opposite faces of the paralGCB, G FB, FB C, and (I. ax. 11.) is lelopiped (12. Cor.). equal to it; and, therefore, adding the Lastly, therefore, let the parallelosolid ABCDH F to each, the prism piped be divided by the plane of any two A B C D E F is equal to the prism A B Copposite edges A'C, a c, into the two DHG, that is, to the prism abcdef. triangular prisms d B A C ac, 6 DAC Lastly, let the sides A C, ac, as also

ac: these prisms shall be equal to one A F, af, be equal, but dissimilar. Make another. For the edges 6 D, d B, being the parallelogram AG similar to af, each of them equal to C A (I. 22.), are

equal to one another; and the dihedral angles at those edges are likewise equal (17. Scholium, 3.) and the sides which contain them are equal, each to each, because they are opposite faces of the parallelopiped. Therefore the prisms are equal to one another (20.).


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Therefore, &c.

tively; that is, to the squares (I. 41. Cor.) Cor. 1. If a triangular prism be com- of the twelve edges, which are made up pleted into a parallelopiped (which may of the sides of these parallelograms, and be done by completing its triangular the four edges before mentioned. bases into parallelograms), the prism shall be equal to half theparallelopiped.

PROP. 23. (Euc. xi. 29 and 30.) Cor. 2. If a pa

Parallelopipeds upon the same base, rallelopiped be cut

and between the same parallel planes, are by two planes which

equal to one another. are drawn parallel respectively to two of

Let ABCD be the common base of its adjoining faces,

two parallelopipeds, which have their the line of their in- A

opposite bases EFGH and KLMN tersection will be parallel to the edge lying in the same plane: the two paralmade by those faces (12. Cor.); and if lelopipeds shall be equal to one another. that line be in the diagonal plane, and

Since E F and KL are both parallel the parallelopiped be thus divided into to A B, either they four parallelopipeds, two of which are parts of the same are about the diagonal plane ; the straight line, or they other two which, together with the for- are parallel to one I

And mer, make up the whole parallelopiped, another (6.), shall be equal to one another. For, as

first let them lie in the in I. 23, the whole parallelopiped, and same straight line. the two parts which are about the di. Then, because LM agonal plane, are bisected by that plane. and E H are equal and parallel to BC,

and therefore to one another (I. ax. I. Scholium.

and 6.), M H is parallel to LE or KF It is worthy of remark, that “the (I. 21.), and MN, G H lie in the same four diagonals of a parallelopiped pass straight line with MH and with each through the same point, and that the other. Therefore, the sides AL, DM, of sum of their squares is equal to the sum

the one parallelopiped, are in the same of the squares of the twelve edges." planes with the sides A F, DG of the For, if the diagonal planes which pass other, each with each. And because the through A C,ac and

bases KM, EG are equal each to the BD, bd intersect

base AC (22.), they are equal to one one another in the

another. Therefore, also, the paralleloline EF, the figures

gram K H is equal to the parallelogram A Cac and B Dbd

LG. And because the triangular prisms will be parallelo

KNHEAD, LMGFBC have the grams, having their opposite sides bi- edges KN, LM equal to one another, and sected by the line E F, (I. 22.); where the dihedral angles at those edges equal fore, the diagonals of each (which are (17. Scholium, 2.), and the sides conthe same with the diagonals of the pa- taining those angles equal (22.), each to rallelopiped), as, for instance, A a, will each, they are equal to one another (21.): bisect EF; because the triangles A EG, but, if the first of these prisms be taken a F G having the angles of the one equal from the whole solid A DNKFGCB, to the angles of the other, each to each there remains the parallelopiped AG; (I. 15.), and the side A E equal to the and if the other be taken from the same side a F, the side EG is likewise equal solid, there remains the parallelopiped to the side FG (I. 5.). Therefore, the AM: therefore the parallelopiped AG four diagonals of the parallelepiped pass

is equal to the parallelopiped A M. (I. through the same point, viz. the middle ax. 3.) point of EF. And again, because A Cac Next, let E F, K L not be in the same and B Dbd are parallelograms, the straight line. Let the planes ADHE, squares of their four diagonals are

BCGF be produced to meet the planes (1.41. Cor.) together equal to the squares ABLK, DCMN, themselves proof A C, a c, A c, aC, and B D, , Bd, duced if necessary, and let the parallelo6 D, that is to say, to the squares of the piped AP be completed* by producing four edges AC, ac, Bd, 6 D, together with the squares of Ac, B D, and aC, completed are already drawn, viz. by producing the

* In this case, the sides of the parallelopiped to be bd, which are the diagonals of the pas plete the parallelopiped, it is only necessary to draw

sides of the parallelopipeds A M, A G*; and, to com. rallelograms ABCD and a b c d respec- the upper base by producing the plane of EFGH and

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the plane of the upper bases EFGH, duce AB to R, so that BR may be KLMN. Then, by the former case, equal to KL; complete the parallelothe parallelopiped AG is equal to AP, gram BCSR; join S B, and produce because OP, FG are in the same to meet D A produced in T; through T

draw T U parallel to AB, to meet CB and S R produced in the points V and U; complete the parallelopiped DX upon the base DTUS, and with the edge DH; produce the planes ABFE, CB FG, in order to complete the parallelopiped BX, and draw the diagonal plane STY Z of the parallelopiped D X. Then the parallelopipeds B H and BX

are equal to one another, because, togestraight line; and, for the like reason, ther with the parallelopipeds BÝ and the parallelopiped AM is equal to the BZ about the diagonal plane, they comsame A P. "Therefore (I. ax. 1.) the plete the whole parallelopiped DX parallelopipeds AG, A M are equal to (22. Cor. 2.). Again, because the bases one another.

BU and BD are complements of the Therefore, &c.

parallelogram DU, BU is equal to

BD (I. 23.), that is, to LN; bút BR PROP. 24. (Euc. xi. 31.) is equal to KL, and the angle R BV

to the angle ABC (I. 3.), that is, to Parallelopipeds upon equal bases, and between the same parallel planes, are

KLM: therefore, also, BV is equal to

L M, and the bases BU, LN are both equal to one another. Let AG, K Q be two parallelopipeds first case, the parallelopiped KQ is

equal and similar. Therefore, by the upon equal bases A B CD, KLMN, equal to the parallelopiped B X, that is, and between the same parallel planes; to BH or AĠ. the parallelopipeds AG, KQ shall be

Lastly, let the bases ABCD,KLMN equal to one another.

be equal, but not equiangular. Make the The bases ABCD, KLMN are either similar, or equiangular, or not equiangular. In the first case,

since they are equal as well as similar, they may be made to coincide; and the parallelopipeds will then stand upon a common base, and between the same parallel planes: therefore they are equal to one another.

Secondly, let them be equiangu- angle ABS, in the plane ABCD, lar, and not similar, having the angle

equal to the angle KLM, and complete at B equal to the angle at L; and the parallelogram ABST, and the patherefore, also, their other angles equal,

rallelopiped XX, upon the base A BST each to each, (1, 15. and I. ax. 3.). Pro

and with the edge BF. Then it is evident that the triangular prisms DH E ATY, CG FBSX have the edges DH, CG equal to one another (22.), and the dihedral angles at those edges equal (17. Scholium 2.), and the sides containing those angles equal, each to each, for the sides A H, B G are opposite faces of the parallelopiped AG (22.), and the sides TÌ, SG stand upon equal

bases (I. ax. 1. and ax. 2. or 3.) TD, SC, KLMN to cut the sides. It may be observed, how. ever, that a parallelopiped can always be completed and between the same parallels (I. 25.); when its base A B C D and one of its principal edges therefore, the prisms are equal to one AE are given, viz. by drawing the side-planes EAB, EAD, and then through BC, CD the side-planes BG, another (21.); and, these being taken from CH parallel to E A D and E A B respectively, and, the whole solid A EYTCGFB, there lastly, through the point E, the upper base EG pat remains the parallelopiped AG, equal rallel to ABCD. This operation is analogons to that by which a parallelogram is coinpleted from two ad. to A X. But the parallelopiped K Q is joining sides (see note, page 1?); and will be some equal to AX by the preceding case; times understood when it is directed to complete a parallelopiped in future propositions.

because their bases are equal (1. 24. and



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