1. Let the solid angle of the polyhedron in question be contained by three angles of equilateral triangles. Upon the given edge AB, describe (I. 42.) an equilateral triangle ABC: take the centre O (III. 26.) of the triangle ABC, and from O draw OD perpendicular to

A

the plane A B C (37.): in OD take the point D such that AD may be equal to AB,* and join DA, DB, DC. Then, because OB and OC are each of them equal to OA, DB and D C are each of them equal to D A or AB (8.), and, therefore, the faces D AB, DAC, DBC of the solid D A B C are equilateral triangles: and because its solid angles are each of them contained by three angles of equilateral triangles, they are equal to one another (49. Cor. 2.). Therefore, D A B C is a regular solid; and since it has four faces, it is a regular tetrahedron.

a

D

E

A

F

C

B

2. Let the solid angle be contained by four angles of equilateral triangles. Upon the given edge A B describe square A B C D (I. 52.); take the centre O (III. 26.), and from O draw OE (37.) perpendicular to the plane ABCD; take OE equal to O A, and, therefore, (I. 22. Cor. 2. and I. 36.) such that AE shall be equal to A B, and produce E O to F, so that O F be equal to OE: join EA, EB, EC, ED, and FA, FB, FC, FD. Then, because the point O is equally distant from the points A, B, C, D, EB, E C, ED are each of them equal to EA or AB (8.): and for the like reason FB, FC, FD are each of them equal to FA; that is, because OF is equal to OE,

may

*That is, take O D2 equal to A B2— A O2 (I, 59.)

to E A, (I. 4.) or AB. Therefore, the eight faces of the solid E A B C D F are equilateral triangles. Again, each of the quadrilaterals EA FC, EBFD is a square, because its diagonals are equal, and (I. 22. Cor. 2.) bisect one another at right angles: therefore, the planes of any two adjoining faces are inclined to one another at the same angle as the planes of any other two, (49. Cor. 2.) viz., the angle which is contained by two angles of equilateral triangles forming with the angle of a square a solid angle. Therefore, any two of the solid incide, and are equal to one another. angles, as E and F, may be made to coTherefore, EABCDF is a regular solid; and, since it has eight faces, it is a regular octahedron.

3. Let the solid angle be contained by

five angles of equilateral triangles. Upon the given edge A B describe a regular pentagon ABCD E(III. 63. Cor.): take the centre O (III. 26.), and from O draw OF perpendicular to the plane ABCDE (37.): take O F such that FA may be equal to A B (which may be done by joining OA, and taking (I. 59.) OF such that the square of OF may be equal to the difference of the squares of O A, A B), and join FA, FB, FC, FD, FE. Then, because the point O is equally distant from the points A, B, C, D, E, FB, FC, FD, FE are equal each of them to FA or AB (8.), and therefore FAB, FBC, FCD, FDE, FEA are equilateral triangles. Join BD; then because the triangles B FD, B C D have the three sides of the one equal to the three sides of the other, each to each, the angle BFD is equal to the angle BCD (I. 7.), that is, to the angle of a regular pentagon; and the same is evidently true of the plane angle formed by any two of the principal edges of the

pyramid FABCDE, which are not adjacent to one another. Complete the regular pentagon BFDGH, and join CG, CH. Then, because the point Cis equally distant from the three points B, F, D, it lies in that perpendicular to the plane BFD, which passes through the centre of the circle circumscribing the triangle B FD (37. Cor.), that is, through the centre of the pentagon BFDGH; therefore CG, CH are each of them equal to CB (8.) or AB, and C D G, CGH, CHB are equilateral triangles. Join CE, and let it cut BD in the point Z. Then, because the diagonal BD is common to the two pentagons A B C D E, H B F D G, and that another diagonal C E of the former cuts it in Z, it may easily be shown that the diagonal FG of the latter cuts it in the same point Z. Therefore EF, FC, and CG lie in one plane; and because the angles EFC, FCG, as shown in the case of the pyramid first constructed, are equal to the angles of a regular pentagon, let the regular pentagon EF CGK be completed, and join DK. Then, as above, it may be shown that DK is equal to DE or AB, and that DEK, DKG are equilateral triangles. Complete in like manner the pentagon AFDK L, and join LE; and in like manner the pentagon B FE L M, and join MA : then, as before, E L is equal to EA or AB, and EAL, ELK are equilateral triangles; and AM is equal to AB, and ABM, A M L are equilateral triangles; and because FA, AM, FC, CH may be shown in the same way as above to be in one plane, and that they make with one another angles of a regular pentagon, the line MH will complete the regular pentagon FAMH C, and B M H will be an equilateral triangle.

Again, because G K is parallel to the diagonal C E of the pentagon FCG KE, and K L to the diagonal D A of the pentagon FDKLA, and so on, the lines GK, K L, LM, MH, H G lie all in the same plane parallel to the plane ABCDE (15.); and the angle contained by every adjacent two is equal to the angle of a pentagon, because KGH is equal to EZB (15.), that is, to E A B (I. 22), and so on. Therefore G K L M H is a regular pentagon. From the centre P of this * Because the diagonals of a regular pentagon divide one another always in the same (viz. the medial) ratio. In fact, the diagonals BD, EC of the pentagon ABCDE being parallel to the sides AE, AB respectively (III. 26. III. 12. Cor. 1. and III. 18.), it follows (1.22.) that EZ is equal to AB or DC; and the triangles ECD, DCZ (I. 6.) being isosceles and similar, EC is to CD (or EZ) as CD (or EZ) to CZ (II. 31.).

pentagon draw (37.) a perpendicular PN' and take P N such (I. 59.) that N G may be equal to GK, and join N G, NK, NL, NM, NH: then NG K, N KL, NLM, NMH, NHG are equilateral triangles. Therefore all the faces of the solid FN are equilateral triangles. And it is evident that any two adjoining faces on the same side of the plane GKLMH are inclined to one another at thesame angle as any other two (49. Cor. 2.), viz., the angle which is contained by two angles of equilateral triangles forming a solid angle with the angle of a pentagon. Also, that any two upon opposite sides of that plane, as A LM, NLM, are inclined to one another at the same angle, may be shown by comparing it with the angle contained by any two G D C, FDC, upon opposite sides of the plane ABCDE: for, since a solid angle is formed at C by the three plane angles BCD, BC F, D C F, of which the first is an angle of a regular pentagon, and the two others angles of equilateral triangles, and in the same manner a solid angle at L by the three plane angles KLM, KLN, M L N, of which the first is an angle of a regular pentagon, and the two others angles of equilateral triangles, the dihedral angle, contained at the edge CD by the planes B CD, FCD, is equal to the dihedral angle contained at the edge L M by the planes K L M, NLM (49. Cor. 2.): and again, since a solid angle is formed at C by the three plane angles BCD, BCG, DC G, of which the two first are angles of regular pentagons, and the other an angle of an equilateral triangle, and in the same manner a solid angle at L by the three angles KLM, KLA, M LÅ, of which the two first are angles of regular pentagons, and the other an angle of an equilateral triangle, the dihedral angle contained at the edge CD by the planes BCD, GCD, is equal to the dihedral angle contained at the edge L M by the planes KLM, ALM (49. Cor. 2.); therefore, the whole dihedral angle contained at the edge CD by the planes FCD, GCD, is equal to the whole dihedral angle contained at the edge L M by the planes N LM, A L M (I. ax. 2.). Therefore, all the plane and likewise all the dihedral angles of the solid FN are equal to one another; and, consequently, any two of its solid angles, as at A and K, may be made to coincide, and are equal to one another. Therefore, FN is a regular solid (def. 9.); and, since it has twenty faces (viz. five, forming the pyramid which has the ver

EC. Then it is evident that the six faces of the parallelopiped EC are squares (I. def. 20. and IV. 22.); and because each of its solid angles is contained by three right angles, any two of them may be made to coincide, and are equal to one another. Therefore EC is a regular solid, and since it has six faces, is a regular hexahedron. This figure is the same as the cube (def. 13.). 5. Let the solid angle be contained by three angles of pentagons. Upon the given edge A B describe a pentagon ABCDE: find the angle, I, at which two angles of regular pentagons must be mutually inclined, in order that with a third angle, which is likewise an angle of a regular pentagon, they may contain a solid angle (49.); and through the sides A B, BC, &c. of the pentagon, already described, draw planes, each making with it that angle of inclination, and intersecting one another in the lines AF, BG, CH, DK, FL: then the angles K

BCHNG, &c.: then, since the angles BGM, BGN are angles of pentagons, and so inclined, that with a third angle of a pentagon they may form a solid angle, the third angle M G N is an angle of a pentagon; and for the like reason the third angles at H, K, L, F are likewise angles of pentagons. Now, it is evident from the figure thus far constructed, that, if at adjacent angles FAB, GBA of a pentagon, other angles FAE, GBC of pentagons be placed at the inclination I, the edges A E, A B, BC will lie in one plane.* If, therefore, the pentagon GMRSN be completed, not only will H N S be an angle of a pentagon, for the reason before stated, but, for the reason just mentioned, OH, HN, NS will lie in one plane, because at the adjacent angles CHN, GNH, other angles CHO, GNS of pentagons are placed at the inclination I: and for the like reasons FMR is an angle of a pentagon, and QF, FM, MR lie in one plane. In like manner, if the pentagon HNSTO be completed, PK, KO, OT will be in one plane; and if KOTU Plie completed, Q L, LP, PU will lie in one plane; and if LPU VQ be completed, M F, F Q, QV will lie in one plane, viz., the plane QFM, or QFMR, and therefore M R being joined will complete the pentagon F Q V R M. Lastly, also, by the same rule, the lines RS, ST, TU, U V, V R lie in one plane, and make with one another angles of pentagons, and therefore RSTUV is a pentagon completing the solid A T. And because AT has all its faces regular pentagons, and all its solid angles (49. Cor. 2.) equal (for each of them is contained by three angles of pentagons), it is a regular solid; and, since the faces are twelve in number, it is a regular dodecahedron.

And, in every case the regular solid is described with the given edge A B. Therefore, &c.

As it is possible that the dotted or occult lines, which have been necessarily introduced in the foregoing constructions, may prevent the reader's acquiring from them a clear notion of the solids intended, we have here added shaded representations of the five regular solids, each in two different positions, in which only so much of the convex surface is exhibited as would present itself to the eye if they were opaque bodies.

perpendicular to the face (37.): let the dihedral angle at A B be bisected by a plane cutting OX in X: X shall be the point in question.

Take O' the centre of the adjoining face, and Y the middle point of AB, and join OY, O' Y, XO', X Y. Then, because OY is perpendicular to AB (III. 3.) X Y is likewise perpendicular to A B (4.); also O'Y is perpendicular to the same A B: therefore OY, XY and O'Y lie in one plane (3. Cor. 1.). And because the plane X YA bisects the dihedral angle O Y A O', the angle XY O is equal to the angle X Y O' (17.): also YO is equal to Y O'; therefore X O' is equal to XO (I.4.) and the angle XO'Y to the angle X OY, that is, to a right angle. But the plane X O'Y is perpendicular to the face which has the centre O' (18.), because it is perpendicular to the line A B, through which that face passes. Therefore (18.) X O' is perpendicular to the face which has the centre O'. Now, because X lies in the perpendicular passing through the centre O, it may easily be shown (4. and I. 4.) that the planes XB C, XCD, &c. make dihedral angles with the plane O A B, equal each to the dihedral angle OYAX; also the dihedral angles of the solid are equal to one another; therefore those planes bisect the dihedral angles of the solid at BC, CD, &c. Hence, as above, it may be shown, that the straight lines drawn from X to the centres of each of the adjoining faces are perpendicular to those faces, and equal each to XO. And because the dihedral angles at AB, BC, &c. are bisected by planes meeting the perpendiculars from the centres of those faces in X, the same may be said of the faces Thereadjoining to them, and so on. fore the straight lines drawn from X to the centres of all the faces are perpendicular to them respectively, and equal each to XO. And hence again, because the centres of the faces are equidistant from their several angles, the point X is likewise equidistant from the several angles of the solid (8.). Therefore X is the common centre of two spheres, one inscribed, and the other circumscribed, as before said. The point X is called the centre of the solid.

Cor. 3. Each of the regular solids of six, eight, twelve, and twenty faces has for every face a face opposite and parallel to it, and the opposite edges of M