have DQF+FQE-DQE=APC+CPB-APB, or DFE= ABC; hence the two symmetrical triangles ABC, DEF are equal in surface.

Scholium. The poles P and Q might lie within triangles ABC, DEF: in which case it would be requisite to add the three triangles DQF, FQE, DQE, together, in order to make up the triangle DEF; and in like manner, to add the three

triangles APC, CPB, APB, together, F in order to make up the triangle ABC in all other respects, the de

D

Q P

monstration and the result would still be the same.

A

B

PROPOSITION XIX. THEOREM.

If the circumferences of two great circles intersect each other on the surface of a hemisphere, the sum of the opposite triangles thus formed, is equivalent to the surface of a lune whose angle is equal to the angle formed by the circles.

Let the circumferences AOB, COD, intersect on the hemisphere OACBD; then will the opposite triangles AOC, BOD, be equal to the lune whose angle is BOD.

For, producing the arcs OB, OD, on the other hemisphere, till they meet in N, the arc OBN will be a semi-circumference, and AOB one also; and taking OB from each, we shall have BN=AO.

A

N

B

Hence,

For a like reason, we have DN=CO, and BD=AC. the two triangles AOC, BDN, have their three sides respectively equal; they are therefore symmetrical; hence they are equal in surface (Prop. XVIII.): but the sum of the triangles BDN, BOD, is equivalent to the lune OBNDO, whose angle is BOD: hence, AOC+BOD is equivalent to the lune whose angle is BOD.

Scholium. It is likewise evident that the two spherical pyramids, which have the triangles AOC, BOD, for bases, are together equivalent to the spherical ungula whose angle is BOD.

PROPOSITION XX. THEOREM.

The surface of a spherical triangle is measured by the excess of the sum of its three angles above two right angles, multiplied by the tri-rectangular triangle.

H

I

G

T

A

B

D

Let ABC be the proposed triangle: produce its sides till they meet the great circle DEFG drawn at pleasure without the triangle. By the last Theorem, the two triangles ADE, AGH, are together equivalent to the lune whose angle is A, and which is measured by 2A.T (Prop. XVII. Cor. 2.). Hence we have ADE+AGH=2A.T; and for a like reason, BGF+BID=2B.T, and CIH+CFE=2C.T But the sum of these six triangles exceeds the hemisphere by twice_the_triangle ABC, and the hemisphere is represented by 4T; therefore, twice the triangle ABC is equal to 2A.T+2B.T+2C.T—4 T; and consequently, once ABC=(A+B+C—2)T; hence every spherical triangle is measured by the sum of all its angles minus two right angles, multiplied by the tri-rectangular triangle.

Cor. 1. However many right angles there may be in the sum of the three angles minus two right angles, just so many tri-rectangular triangles, or eighths of the sphere, will the proposed triangle contain. If the angles, for example, are each equal to of a right angle, the three angles will amount to 4 right angles, and the sum of the angles minus two right angles will be represented by 4-2 or 2; therefore the surface of the triangle will be equal to two tri-rectangular triangles, or to the fourth part of the whole surface of the sphere.

Scholium. While the spherical triangle ABC is compared with the tri-rectangular triangle, the spherical pyramid, which has ABC for its base, is compared with the tri-rectangular pyramid, and a similar proportion is found to subsist between them. The solid angle at the vertex of the pyramid, is in like manner compared with the solid angle at the vertex of the trirectangular pyramid. These comparisons are founded on the coincidence of the corresponding parts. If the bases of the

pyramids coincide, the pyramids themselves will evidently coincide, and likewise the solid angles at their vertices. From this, some consequences are deduced.

First. Two triangular spherical pyramids are to each other as their bases: and since a polygonal pyramid may always be divided into a certain number of triangular ones, it follows that any two spherical pyramids are to each other, as the polygons which form their bases.

Second. The solid angles at the vertices of these pyramids, are also as their bases; hence, for comparing any two solid angles, we have merely to place their vertices at the centres of two equal spheres, and the solid angles will be to each other as the spherical polygons intercepted between their planes or faces.

The vertical angle of the tri-rectangular pyramid is formed by three planes at right angles to each other: this angle, which may be called a right solid angle, will serve as a very natural unit of measure for all other solid angles. If, for example, the the area of the triangle is of the tri-rectangular triangle, then the corresponding solid angle will also be of the right solid angle.

PROPOSITION XXI. THEOREM.

The surface of a spherical polygon is measured by the sum of all its angles,minus two right angles multiplied by the number of sides in the polygon less two, into the tri-rectangular triangle.

D

B

From one of the vertices A, let diagonals AC, AD be drawn to all the other vertices; the polygon ABCDE will be divided into as many triangles minus two as E it has sides. But the surface of each triangle is measured by the sum of all its angles minus two right angles, into the trirectangular triangle; and the sum of the angles in all the triangles is evidently the same as that of all the angles of the polygon; hence, the surface of the polygon is equal to the sum of all its angles, diminished by twice as many right angles as it has sides less two, into the tri-rectangular triangle.

A

Scholium. Let s be the sum of all the angles in a spherical polygon, n the number of its sides, and T the tri-rectangular triangle; the right angle being taken for unity, the surface of the polygon will be measured by

(s—2 (n—2,)) T, or (s—2 n+4) T

APPENDIX.

THE REGULAR POLYEDRONS.

A regular polyedron is one whose faces are all equal regular polygons, and whose solid angles are all equal to each other. There are five such polyedrons.

First. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three of those triangles, by four, or by five: hence arise three regular bodies, the tetraedron, the octaedron, the icosaedron. No other can be formed with equilateral triangles; for six angles of such a triangle are equal to four right angles, and cannot form a solid angle (Book VI. Prop. XX.).

Secondly. If the faces are squares, their angles may be arranged by threes: hence results the hexaedron or cube. Four angles of a square are equal to four right angles, and cannot form a solid angle.

Thirdly. In fine, if the faces are regular pentagons, their angles likewise may be arranged by threes: the regular dode caedron will result.

We can proceed no farther: three angles of a regular hexagon are equal to four right angles; three of a heptagon are greater.

Hence there can only be five regular polyedrons; three formed with equilateral triangles, one with squares, and one with pentagons.

Construction of the Tetraedron.

Let ABC be the equilateral triangle which is to form one face of the tetraedron. At the point O, the centre of this triangle, erect OS perpendicular to the A planc ABC; terminate this perpendicular in S, so that AS-AB; draw SB, SC: the pyramid S-ABC will be the tetraedron required.

For, by reason of the equal distances

S

O

B

OA, OB, OC, the oblique lines SA, SB, SC, are equally re

moved from the perpendicular SO, and
consequently equal (Book VI. Prop. V.).
One of them SA=AB; hence the four
faces of the pyramid S-ABC, are trian- A
gles, equal to the given triangle ABC.
And the solid angles of this pyramid
are all equal, because each of them is
formed by three equal plane angles:
hence this pyramid is a regular tetrae-
dron.

Construction of the Hexaedron.

Let ABCD be a given square. On the base ABCD, construct a right prism whose altitude AE shall be equal to the side AB. The faces of this prism will evidently be equal squares; and its solid angles all equal, each being formed with three right angles : hence this prism is a regular hexaedron or cube.

B

H

D

E

A

The following propositions can be easily proved.

G

B

1. Any regular polyedron may be divided into as many regular pyramids as the polyedron has faces; the common vertex of these pyramids will be the centre of the polyedron; and at the same time, that of the inscribed and of the circumscribed sphere.

2. The solidity of a regular polyedron is equal to its surface multiplied by a third part of the radius of the inscribed sphere.

3. Two regular polyedrons of the same name, are two similar solids, and their homologous dimensions are proportional; hence the radii of the inscribed or the circumscribed spheres are to each other as the sides of the polyedrons.

4. If a regular polyedron is inscribed in a sphere, the planes drawn from the centre, through the different edges, will divide the surface of the sphere into as many spherical polygons, all equal and similar, as the polyedron has faces.