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PROP. XVIIT. THEOR.

The plain angles, which constitute any solid angle, are together less

than four right angles.

Let A be a solid angle, contained by any number of plain angles BAC, CAD, DAE, EAF and FAB, these are together less

B than four right angles.

Let the plains which contain the solid angle A be cut by another plain BCDEF; and of the three plain angles which contain

E the solid angle at B, the angles ABF and ABC are together greater than the third CBF (17.2 Sup.); for the same reason, of the three plain angles which contain each of the solid angles at C, D, E and F, the two which are at the bases of the triangles having their common vertex at A, are together greater than the third, which is one of the angles of the figure BCDEF; therefore all the angles at the bases of the triangles having their common vertex at A are together greater than all the angles of the figure BCDEF; but all the angles of the figure BCDEF are equal to twice as many right angles, except four, as the figure has sides (Cor. 1. 32. i Eu.), therefore all the angles at the bases of the triangles having their common vertex at A are greater than twice as many right angles, except four, as the figure has sides, and all the angles of these triangles are equal to twice as many right angles as the figure has sides (32. 1 Eu.), therefore the angles of these triangles which are at their common vertex A, being those which contain the solid angle A, are less than four right angles :

Cor. From this proposition it follows, that there can be no more than five solids contained by equilateral and equiangular plain figures, or as they are usually called, regular solids, namely, three contained by equilateral triangles, one by squares, and one by regular pentagons.

For a solid angle cannot be contained by two plain angles, three at least are required.

And since the three angles of an equilateral triangle are equal to two right angles (32. 1 E.), six such angles are equal to four right angles, and therefore cannot constitute a solid angle (by this prop.); and since six angles of an equilateral triangle are equal to four right angles, three, four, or five such angles are less than four right angles, and can therefore constitute a solid angle, as is manifest from this proposition; but three such angles form the angle of a tetrahedron or equilateral pyramid, see Def. 31. 2 Sup. ; four such angles form the angle of an octohedron, see Def. 33. 2 Sup. ; and five such angles form the angle of an icosihedron, see Def. 35. 2 Sup.

Three angles of a square form the angle of a cube or hexahedron, see Def. 32. 2 Sup. ; four such angles are equal to four right angles, and therefore cannot constitute a solid angle.

And since the five angles of a regular pentagon are equal to six right angles (Cor. 1. 32. 1 Eu.), any one of its angles is equal to a right angle and a fifth of a right angle, and therefore three such angles are equal to three right angles and three fifths, and three such angles form the angle of a dodecahedron, see Def. 34. 2. Sup. ; but four such angles are equal to four right angles and four fifths, and therefore cannot form a solid angle (by this prop).

And since six angles of an equilateral triangle, or four angles of a square, are equal to four right angles, and four angles of a regular pentagon are greater than four right angles, therefore more than six angles of an equilateral triangle, or than four of a square, or than four of a regular pentagon, are greater than four right angles, and therefore cannot constitute a solid angle (by this prop).

And since the six angles of a regular hexagon are equal to eight right angles (Cor. 1. 32. 1 Eu.), three such angles are equal to four right angles, and therefore cannot constitute a solid angle (by this prop); neither therefore can any greater number.

And since three angles of a regular hexagon are equal to four right angles, three angles of a regular heptagon, or of any regular polygon of more than six sides, are greater than four angles, as also easily follows from Cor. 1. 32. 1 Eu., therefore all regular polygons of more than five sides are incapable of forming a solid angle, and therefore there can be no more regular solids than the five mentioned in this corollary.

Schol.-It is manifestly supposed in this proposition, that when the solid angle is contained by more than three plain angles, any of the legs of the plain angles which form the solid angle, as AČ, falls without the plain passing by the two adjacent legs AB and AD.

PROP. XIX. THEOR.

The section of a plain with the surface of a sphere (ADBE), is :

circle, whose centre is in the diameter of the sphere which is perpendicular to the plain. Part 1. If the section pass

D through the centre of the

L sphere, as the section AFBG, H the proposition is manifest,

K since all right lines, as CB and

I

M CG, drawn from the centre C to the surface of the sphere, and A therefore to the perimeter of the section, are equal (Def. 12.

G 2 Sup.), which section is therefore a circle (Def. 10.1 Eu).

Part 2. But if the section do not pass through the centre of the sphere, as the section HLKM, Iet CO be a perpendicular drawn from the centre C# the plain HLKM (6. 2 Sup.); through O draw, in the plain ILK, any two right lines whatever ÉK and LM, meeting the section HLKM in the points H and K, L and M, and draw CM and Ck; and because the triangles COK and COM have the angles at 0 right (Constr. and Def. 3. 2 Sup.), the sides CK and CM equal to each other (Def. 12. 2 Sup.), and CO common to both, the sides OK and OM are equal (Cor: 7. 6 Eu.); in like manner, may all right lines drawn from 0 to the perimeter of the section HLKM be proved equal to each other, which section is therefore a circle (Def. 10. 1 tu.), whose centre is in the diameter of the sphere DE, which is perpendicular to the plain HLKM.

Cor. From this proposition it appears, how the centre of a given sphere may be found, namely, by finding the centre 0 of any circle HLK-formed by the intersection of a plain with its surface (1. 3 Eu.), drawing through that centre a perpendicular to the plain (7. 2 Sup.), to meet the surface of the sphere both ways as in D and E, and bisecting DE in C; the point C is the centre of the sphere.

If the centre of the sphere be in DE, 'tis plain, C must be that centre; if the centre were in any point without DE, right lines being drawn, in the plain passing by DE and that point from that point to O and the intersections of the same plain with the circle HLK, a like absurdity might be shewn to follow, as in 1.

PROP. XX. THEOR.

Any plain, passing through tlie vertex of a conè, and cutting the

circumference of its base, cuts the opposite surfaces in two right lines, and in them only.

For if from the points in which this plain cuts the circumference of the base, two right lines be drawn to the vertex, they are in the cutting plain (Def. 4 and 7. 1 Eu.), and in the conical surface (Def 17.2 Sup.), and, being produced beyond the vertex, in the opposite surface (Def. 22. 2 Sup.); therefore the plain passing through the vertex cuts the opposite surfaces in these two right lines ; and since the intersection of this plain with the base, which is a right line (1. 2 Sup.), cannot cut the circumference of the base in more than two points, that plain cannot cut the coni-. cal surface in more than these two right lines.

Cor. The intersection of a plain, passing through the vertex of a cone, with the conical surface and base, is a triangle.

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If either of the opposite surfaces of a cone be cut by a plain parallel

to the base, the section is the circumference of a circle whose centre is in the axis of the cone.

Let ABD be a cone, whose vertex is A, and its base the circle BKDL, of which C is the centre ; AC is its axis (Def. 20. 2 Sup.) ; let EGFH be the section of a plain parallel to the base, with one of the opposite surfaces; EGFH is a circle, whose centre is in the axis AOC.

A Let the cone be cut by any two plains passing by the axis AC, mining the triangles ABD and AKL vor. 20. 2 Sup.), and let these trianzies, produced if necessary, meet the pain EGFH in the right

S lines Eur and GOH ; because of the par

H allel plains, the right lines EF and BD, as also GH and KL, are parallel (15. 2 K Sup.); therefore the triangles AOF and

BE ACD, as also AOH and ACL, are equiangular, and therefore the ratios of CD to OF, and of CL to OH, are each of them equal to that of AC to AO (4. 6 and 16.5 Eu.), and therefore to each other (11. 5 Eu.); whence, CD and CL being, on account of the circular base of the cone, equal, OF and OH are equal (14. 5 Eu). In like manner it may be shewn, that any other two right lines, drawn from 0 to the section EGFH are equal; since therefore all right lines drawn from 0 to that section are equal, that section is the circumference of a circle whose centre is 0, and therefore in the axis AC of the cone.

Cor. Hence it follows, that any diameter of such a section passes through the axis of the cone; and any right line, drawn in the plain of such a section, through the axis of the cone, is a: diameter of the section.

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