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triangle A B C, which is the case also with the right angle of a right-angled plane triangle (I. 19. Cor. 3.).

Scholium.

In a triangle A B C which has one of its angles A B C equal to the sum of the other two, the containing sides AB, BC are together less than a semicircumference. For G, the middle point of AC, is the pole of a small circle passing through A, B, and C, because G A, G B and G C are equal to one another: and, because a spherical are (9. Schol.) is the shortest distance between two points along the surface of the sphere, the great arcs AB and BC are together less than the small arcs AB and B C, that is, than the semi-circumference of a small circle: much more, therefore, are AB and BC together less than the semicircumference of a great circle. In fact, when the given sides A B and B C are together equal to a semicircumference, BC coincides with BR, the polar diameter of the circle BCK, and because AC is always bisected by the circumference of the great circle HGN, AG and GC are in this case quadrants, and consequently (19. Cor. 1.) AGC is at right angles to A B; so that ABC is a lune, not a triangle, and equal to a fourth part of the surface of the sphere. And when the given sides A B and B C are together greater than a semicircumference, it is evident, without reference to the figure, that a triangle may be found which shall have AB, BC for two of its sides, and differ by as little as we please from half the surface of the sphere. For, in this case, if A B, B C are placed in the circumference of the same great circle, that is, at an angle equal to two right angles, the arc AC which completes the circle will be less than a semicircumference; and if A B, BC are placed at any angle less than two right angles, the arc AC will be still less (18.); therefore, if A B, BC are placed at any angle less than two right angles, A C being joined will complete a spherical triangle (def. 9.), and this triangle will differ less and less from the hemispherical surface, as the angle A B C approaches to two right angles.

In a triangle ABC of this kind, the angle ABC, which is equal to the sum of the other two, is always greater than a right angle, because the three together are greater than two right angles (10). It is worthy of notice, also, that the triangle B'A C, which is the difference between

the lune BA B' C, and the triangle ABC has its surface always equal to a fourth part of the surface of the sphere; for it is measured by the difference of 2 B and 2 B-2 R, that is, by 2 R, and the sphere's surface by 8 R (21. Cor. 1.).

It may be observed that a similar course of reasoning to that which is exhibited in Book III. Prop. 40, 41, 42, and 43, founded upon this proposition, will lead to the same conclusions with regard to the surfaces included by spherical polygons and circles of the sphere, which are there stated with regard to the areas of rectilineal figures, and circles on a plane surface: viz.

1. Of all spherical polygons, contained by the same given sides, that one contains the greatest portion of spherical surface, which has all its angles in the circumference of a circle.

2. A circle includes a greater portion of the spherical surface than any spherical polygon of the same perimeter.

3. The lunular surface, which is included by a spherical arc and a small arc, is greater than any other surface which is included by the same perimeter, of which the same spherical arc is a part.

A

We may also infer from Prop. 25., that, of all spherical polygons having the same number of sides and the same perimeter, the greatest is that which has all its sides equal and all its angles equal. For if a spherical polygon ABCDE have not all its sides equal, and A B, A E be two adjoining sides which are unequal, a greater ABCDE may be found with the same perimeter by describing upon the base B E the isosceles triangle A'BE, which has the same perimeter with ABE. And, as above, B if a spherical polygon have not all its angles lying in the circumference of a circle, a greater may be found with the same sides. Therefore, none is greatest, but that which has all its sides equal, and all its angles lying in the circumference of a circle, that is, which has all its sides equal and all its angles equal; and, since there is evidently some greatest, the greatest is that which has all its sides equal and all its angles equal.

PROP. 27.

C

E

Spherical pyramids, which stand upon equal bases, are equal to one another; so, likewise, are their solid angles.

First, let the bases of the pyramids be equal triangles, which have one side

of the one equal to one side of the other, and let the equal sides be made to coincide, so that the triangles A B C, EBC (in the figure of Prop. 24.) may represent the bases of the pyramids. Then the triangles ABC, EB C, being equal to one another, lie between the same equal and parallel small circles, and may be completed (as in Prop. 24.) into the quadrilaterals A B C D, EBC F. And it may be shown, as in the same proposition, that the triangles EAB, FDC have the three sides of the one equal to the three sides of the other, each to each, and therefore may be made to coincide (15.): wherefore, also, the pyramids, which have these triangles for their bases, may be made to coincide, and are equal to one another. And hence, as in Prop. 24., it was demonstrated, by the addition and subtraction of the lunular portions of surface, that the quadrilateral A B C D is equal to the quadrilateral EBCF, so, here, it may be demonstrated, by the addition and subtraction of the pyramidal solids (23.), which have these lunular portions for their bases, that the pyramidal solids, which have for their bases the quadrilaterals A B CD and E B C F, are equal to one another. But, because the triangles A B C, CDA have the three sides of the one equal to the three sides of the other, each to each, in the same order, they may be made to coincide (15.), and therefore the pyramids which have these triangles for their bases may be made to coincide, and are equal to one another; and each of them is the half of the pyramidal solid which has the quadrilateral A B CD for its base. And in the same manner it may be shown, that each of the pyramids on the triangular bases, EBC, CFE is the half of the pyramidal solid on the quadrilateral base EBC F. Therefore, because the halves of equals are equal, the pyramid upon the base ABC is equal to the pyramid upon the base EBC.

Next, let the bases of the pyramids be any equal triangles A B C, K L M.

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be taken equal to the triangle A B C or KLM, and join K' M. Then, by the former case, the pyramids upon the bases A B C, K' L M' are equal to one another. And, because the triangle K'L M' is equal to the triangle K L M, the triangles K' M M and K K' M which have the common side K' M are likewise equal; therefore, by the first case, the pyramids, which have these triangles for their bases, are equal to one another; and, these being added to, or taken from, the pyramid upon the base KLM, the whole or remaining pyramid upon the base K'L M' is equal to the whole or remaining pyramid upon the base K L M, that is to the pyramid upon the base A B C.

Lastly, let the bases of the pyramids P, P' be equal polygons. Let a triangle be found which is equal to one, and therefore also to the other of the polygons, and let Q be the pyramid which has this triangle for its base. Then, because this triangle is equal to the base of the pyramid P, it may be divided into triangles which are equal respectively to the triangles into which the base of P is divisible. And it has been already shown that pyramids which have equal triangles for their bases are equal to one another; and the sums of equals are equal; therefore, the pyramids P and Q are equal to one another. In the same manner it may be shown that the pyramids P' and Q are equal to one another. Therefore P is equal to P'. And it is evident that what has been shown with regard to spherical pyramids, being derived from coincidence, may be shown equally of their solid angles at the centre of the sphere.

Therefore, &c.

PROP. 28.

Any two spherical pyramids are to one another as their bases; and the solid angles of the pyramids are to one another in the same ratio.

Let P, P' be any two spherical pyramids, and let B, B' be their bases: the pyramid P shall be to the pyramid P/ as the base B to the base B'.

For if the base B' be divided into any number of equal parts, then, because pyramids which stand upon equal bases are equal to one another, the pyramid Pr will be divided into the same number of equal parts by planes passing through the arcs of division (27.); and if the base B contain exactly, or with a remainder,` a certain number of parts equal to the

former, the pyramid P will contain exactly, or with a remainder, the same number of parts equal to the latter (27.): therefore, P is to P' as B to B' (II. def. 7.)

And the same may be said of the solid angles.

Therefore, &c.

Cor. Every spherical pyramid is equal to the third part of the product of its base and the radius of the sphere.

For, if the whole sphere be divided into spherical pyramids, these pyramids will be to one another as their bases; and, therefore, any one of them is to the sum of all, as the base of that one to the sum of all the bases (II. 25. Cor. 3.); that is, any pyramid is to the whole sphere as its base to the surface of the sphere. Therefore, any pyramid is to the whole sphere as the product of its base and the radius of the sphere to the product of the whole surface and radius (IV. 26.); and because, in this proportion, the second term is equal to the third part of the fourth (V. 17.), the first term is likewise equal to the third part of the third (II. 19. and II. 13.).

Scholium.

Hence it appears that every solid angle is measured by the spherical surface which is described with a given radius about the angular point, and intercepted between its planes. For it is shown in the proposition that any two solid angles are to one another as these surfaces.

This measure bears an obvious analogy to the measure of a plane angle, as stated in the Scholium at Book III. Prop. 13. The angular unit of the latter measurement is the right angle; into four of which the whole angular space about any point in a plane is divided by two straight lines drawn at right angles to one another; and each right angle is measured by a quadrant, or fourth part of the circumference of a circle described about that point with a given radius. In like manner, if, through a point in space, three planes be made to pass at right angles to one another, they will divide the whole angular space about that point into eight solid right angles, each of which (a solid angular unit) is measured by an octant, or eighth part of the surface of a sphere described about that point with a given radius.

Thus, then, the plane angles, dihedral angles, and magnitude of a solid angle, which has three or a greater number of

faces, are represented, upon the surface a sphere described about the angular point, by the sides, angles, and surface of a spherical triangle or polygon: and whatever has been stated with regard to the latter may be understood likewise of the former. It is shown (for instance) in Prop. 13. that if two solid angles, each of which is contained by three plane angles, have two plane angles of the one each to each, and likewise the dihedral equal to two plane angles of the other, angles contained by them equal to one another, the remaining plane and dihedral angles of the one shall be equal to of the other, each to each. And, genethe remaining plane and dihedral angles rally, all questions which relate to solid angles will be placed in the clearest view before us, when we contemplate only their representation and that of their parts, upon the surface of the sphere.

SECTION 4.-Problems.

In the solution of the following problems, it is assumed,

1. That of any two points, which are direct distance may be obtained; and, given upon the surface of a sphere, the

2. That from any given point as a pole, with any given distance less than a semicircumference, a circle may be described upon the surface of a sphere.

We may observe, however, that the first assumption is only used in Prob. 1 to obtain the diameter of the sphere; and, therefore, if the diameter is supposed to be given, it may be dismissed as unnecessary. Constructions made within the sphere are excluded.

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equal to the direct distances B C, CD, B'D respectively; find e the centre of the circumscribing circle (III. 59.), and join e b; from the point e draw ea perpendicular to eb, and from the centre b, with a radius equal to the direct distance B A describe a circle cutting ea in a; produce a e to f, and from b draw bf perpendicular to ba (I. 44.) to meet af in f: af shall be equal to the diameter of the sphere.

For, if O be the centre of the sphere, and the diameter AOF be drawn cutting the plane BCD in E, and EB be joined, A F will be the axis (def. 3.), and therefore (1.) the point E the centre, of the circle B C D, and the angle AEB will be a right angle. Also, EB is equal to e b, because the triangles B CD, bed may be made to coincide, and, then, the points E, e being, each of them, the centre of the same circumscribing circle, will likewise coincide. Therefore, because in the right-angled triangles AEB, aeb, the hypotenuse A B is equal to the hypotenuse ab, and the side EB to the side e b, the angle B A E is equal to the angle bae (I. 13.). Join AB, BF, OB. Then, because O B is equal to the half of A F, the angle ABF is a right angle (I. 19. Cor. 4.); and, because, in the right-angled triangles ABF, abf, the side AB and angle BAF are equal respectively to the side ab and angle baf, the hypotenuses AF and af are equal to one another (I. 5.), and af is equal to the diameter of the sphere.

Therefore, &c.

Cor. Hence, a sphere being given, the quadrant of a great circle may be found; for it is equal to the quadrant of the circle which is described upon the diameter af.

PROP. 30. Prob. 2.

Any point A being given upon the surface of a sphere, to find the opposite extremity of the diameter which passes through that point.

From the pole A, with the distance of a quadrant, (29. Cor.) describe a spherical arc PQ (5.); and, from any two points P, Q of

this arc, as poles, with the same distance, describe two

P

great circles passing through A, and cutting one another again in the point a.

Then, because the planes of any two great circles cut one another in a diameter of the sphere, the points A, a are opposite extremities of a diameter. Therefore, &c.

PROP. 31. Prob. 3.

To join two given points A, B upon the surface of a sphere. From the pole A, with the distance of a quadrant (29. Cor.), describe a great circle, and from the pole B, with the same distance, describe a great circle cut

ing the former in the point P. From P as a pole, with the same distance, describe the arc A B. Then, because AB is the arc of a great circle (5.) described between the points A, B, it joins those two points on the surface of the sphere (def. 5.).

Therefore, &c.

Cor. In the same manner, any spherical arc being given, the great circle may be completed of which it is a part. PROP. 32. Prob. 4.

To bisect a given spherical arc A B. From the pole A, with the distance A B describe a circle, and from the pole B with the same distance AB or BA describe a circle cutting the former circle in the points C, D. Join CD(31.) and let it cut AB in E. AB is bisected in E.

See Book I. Prop. 43.
Therefore, &c.

A

B

E

A

It must here be observed that although the points C, D are determined by the intersection of small circles, the arcs CA, CB, CD, DA, DB, which are the sides of the spherical triangles in the demonstration, are portions of great circles; and the same remark applies to some subsequent problems.

Cor. In the same manner, a spherical arc CD may be drawn which shall bisect any spherical arc A B at right angles.

PROP. 33. Prob. 5.

To draw an arc, which shall be perpendicular to a given spherical arc AB, from a given point C in the same.

From C B, or CB produced, cut off CP equal to a quadrant, and from the pole P,

D

A

B

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To describe a circle through three given points A, B, C, upon the surface of a sphere.

Join (31.) AB and AC, and bisect them at right angles (32. Cor.) with the arcs DP, EP, which meet one another in the points P, P'. From either of these points, P, as a pole, with the distance PA, describe

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D

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a circle. It shall pass through the other two points B and C. For PB is equal to PA, because the triangles PD B, PDA have two sides of the one equal to two sides of the other, each to each, and the included angles PDA, PDB equal to one another (13.); and in the same manner it may be shown that PC, likewise, is equal to PA.

Therefore, &c.

Cor. In the same manner, the poles

To bisect a given spherical angle P, P' of any given circle A B C may be

ВАС.

A

In A B take any point B; make AC equal to A B, and join BC; from the poles B, C, with the common distance B C, describe two spherical arcs cutting one another in D, and join AD (31.). AD is the bisecting arc required. See Book I. Prop. 46. Therefore, &c.

PROP. 36. Prob. 8.

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found.

PROP. 38. Prob. 10.

Through two given points A, B, and a third point C on the surface of a sphere, to describe two equal and parallel small circles; the points A, B, C not lying in the circumference of the same great circle.

Join AB, and draw DE bisecting it at right angles in the point D (32. Cor.); find A' the opposite extremity of the diameter which passes through A (30.); join A' C (31.), and bisect it at right angles with the arc FP which meets DE in the points P, P' (32. Cor.); join PA, PC, and from the pole P with

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BC

D

D, E; and from the pole A, with the same distance, describe a circle BFG cutting A B in B; from B F G cut off BF equal to DE, and join AF (31.). The angle B AF is equal to the given angle C; for they are measured by equal arcs BF, DE (6.) Therefore, &c.

P

the distances PA, PC describe two small circles AGH and CKL: the former shall likewise pass through the point B, and they shall be the two equal and parallel small circles required.

For, if PA and PB be joined, they will be equal to one another, because

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