arcs PA, PB, PC (P. 3); at the point F make the angle DFQ= ACP, the arc FQ-CP; and draw DQ, EQ.

The sides DF, FQ, are equal to the sides AC, CP; the angle DFQ=ACP; hence, the two triangles DFQ, ACP, are equal in

D

Q P

E

B

all their parts (P. 8); consequently, the side_DQ=AP, and the angle DQF=APC.

In the triangles DFE, ABC, the angles DFE, ACB, opposite to the equal sides DE, AB, are equal (P. 10). If the angles DFQ, ACP, which are equal by construction, be taken away from them, there will remain the angle QFE, equal to PCB. The sides QF, FE, are equal to the sides PC, CB; hence, the two triangles FQE, CPB, are equal in all their parts (P. 8); hence, the side QE=PB, and the angle FQE=CPB.

Now, the triangles DFQ, ACP, which have their sides respectively equal, are at the same time isosceles, and capable of coinciding, when applied the one to the other. For, having placed AC on its equal DF, the equal sides will fall the one on the other, and thus the two triangles will exactly coincide: hence, they are equal; and the surface DQF APC. For a like reason, the surface FQE= CPB, and the surface DQE=APB; hence we have,

or,

=

DQF+FQE-DQE=~APC+CPB−APB,
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, in order to make up the triangle ABC: in all other respects, the demonstration and the result would be the same.

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

Let the circumferences A OB, COD, intersect on the surface of a hemisphere; then will the opposite triangles AOC, BOD, be equivalent to the lune whose angle is BOD. For, produce the arcs OB, OD, on

the other hemisphere, till they meet in N. Now, since AOB and OBN are semicircumferences, if we take away the common part OB, we shall have BN AO. For a like reason, we have DN=CO, and BD=AC. Hence, the two triangles AOC, BDN,

B

have their three sides respectively equal: they are therefore symmetrical; hence, they are equal in surface (P. 16). 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.

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

Let ABC be any spherical triangle: then will its sur face be equal to

(A+B+C−2)× T.

For, 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 2AXT (P. 15, c. 2). Hence, we have ADE+ AGH=2AXT; and, for a like reason, BGF+BID=2BXT, and CIH+CFE =20XT. But the sum of these six triangles exceeds the hemisphere by

twice the triangle ABC, and the hemisphere is represented by 47: therefore, twice the triangle ABC, is equivalent to 2AXT+2BXT+20×T-4T;

and, consequently,

ABC≈(A+B+C−2)×T;

hence, every spherical triangle is measured by the sum of its three angles minus two right angles, multiplied by the tri-rectangular triangle.

Scholium 1. When we speak of the spherical angles, we regard the right angle as unity, and compare the sum of the three angles with this standard. Hence, however many right angles there may be in the sum of the three angles minus two right angles, just so many tri-rectangular triangles, will the proposed triangle contain. If the angles, for example, are each equal to of a right angle, the sum of the three angles is equal to 4 right angles; and this sum, minus two right angles, is represented by 4-2, or 2; therefore, the surface of the triangle is equal to two tri-rectangular triangles, or to the fourth part of the surface of the entire sphere.

Scholium 2. The same proportion which exists between the spherical triangle ABC, and the tri-rectangular triangle, exists also between the spherical pyramid which has ABC for its base, and the tri-rectangular pyramid. The triedral angle of the pyramid is to the triedral angle of the trirectangular pyramid, as the triangle ABC to the tri-rectangular triangle. From these relations, the following conse quences 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 pyramids, it follows that any two spherical pyramids are to each other, as the polygons which form their bases.

Second The polyedral angles at the vertices of these pyramids, are also as their bases; hence, for comparing any two polyedral angles, we have merely to place their vertices at the centres of two equal spheres; the angles are to each other as the spherical polygons intercepted between their 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 polyedral angle, will serve as a very natural unit of measure for all other poly edral angles. If, for example, the area of the triangle is of the tri-rectangular triangle, the corresponding trie dral angle is also of the right polyedral angle.

PROPOSITION XIX. THEOREM.

The surface of a spherical polygon is equal to the excess of the sum of all its angles, over two right angles taken as many times as there are sides in the polygon less two, multiplied by the tri-rectangular triangle.

Let ABCDE be a spherical polygon. From one of the vertices A, let diagonals AC, AD, be drawn to the other vertices; the polygon ABCDE will be divided into as many triangles less two, as it has sides.

Now, the surface of each triangle

E

D

A

is equal to the sum of all its angles less two right angles, into the tri-rectangular triangle. The sum of the angles of all the triangles is 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-rectangu lar triangle.

Scholium. Let s be the sum of all the angles of a spheri cal polygon, n the number of its sides, and T the tri-rect angular triangle; the right angle being taken as unity, the surface of the polygon will be equal to

(s—2 (n−2,))×T=(s—2n+4) × T.

APPENDIX.

NOTE A.-PAGE 22.

A DEMONSTRATION is a train of logical arguments brought to a conclusion. a conclusion. The bases or premises of a demonstration, are definitions, axioms, propositions previously established, and hypotheses. The arguments are the links which connect the premises, logically, with the conclusion or ultimate truth to be proved.

In Geometry we employ two kinds of demonstrationthe Direct, and the Indirect or the method involving the Reductio ad absurdum.

These are also called Positive and Negative Demonstra tions. In the direct method, the premises are definitions, axioms, and previous propositions; and by a process of logical argumentation, the magnitudes of which something is to be proved, are shown to bear the mark by which that may always be inferred, or, in other words, are shown to fall under some definition, axiom, or proposition, previously laid down. The direct demonstration may be

divided into two classes:

1st. Where the argument depends on superpositionthat is, on the coincidence of magnitudes when applied the one to the other: and

2dly. Where it depends on addition and subtraction, or immediately on principles previously laid down.

The indirect method rests on a hypothesis. This hypothesis is combined in a process of logical argumentation, with definitions, axioms, and previous propositions, until a conclusion is obtained, which agrees or disagrees with some known truth. Now, if the conclusion so deduced, is excluded from the truths previously established, that is, if