PROPOSITION XV. THEOREM.

Any lune, is to the surface of the sphere, as the arc which measures its angle is to the circumference of a great circle; or, as the angle of the lune is to four right angles.

Let AMBN be a lune, and MCN the angle of the lune, then will the area of the lune be to the surface of the sphere, as the arc MN is to the circumference of a great circle MNPQ; or, as the angle MCN is to four right angles (B. III., P. XVII., C. 2).

In the first place, suppose the arc MN and the circumference MNPQ to be commensurable. For example, let them be to each other as 5 is to 48. Divide the circumference MNPQ into 48 equal parts, beginning at M; MN will contain five of these parts. Join each point

M

of division with the points A and B, by a quadrant: there will be formed 96 equal isosceles spherical triangles (P. VII., S. 2) on the surface of the sphere, of which the lune will contain 10: hence, in this case, the area of the lune is to the surface of the sphere, as 10 is to 96, or as 5 is to 48; that is, as the arc MN is to the circumference MNPQ, or as the angle of the lune is to fo right angles.

In like manner, the same relation may be shown to exist when the arc MN, and the circumference MNPQ are to each other as any other whole numbers.

If the arc MN, and the circumference MNPQ, are not commensurable, the same relation may be shown to exist by

a course of reasoning entirely analogous to that employed in Book IV., Proposition III. Hence, in all cases, the area of a lune is to the surface of the sphere, as the arc measuring the angle is to the circumference of a great circle; or, as the angle of the lune is to four right angles; which was to be proved.

Cor. 1. Lunes, on the same or on equal spheres, are to each other as their angles.

Cor. 2. If we denote the area of a tri-rectangular triangle by T, the area of a lune by L, and the angle of the the right angle being denoted by 1, we shall

lune by A,

have,

whence,

L : 8T :: A: 4;

LTX 2A;

hence, the area of a lune is equal to the area of a trirectangular triangle multiplied by twice the angle of the lune.

Scholium. The spherical wedge, whose angle is MCN, is to the entire sphere, as the angle of the wedge is to four right angles, as may be shown by a course of reasoning entirely analogous to that just employed: hence, we infer that the volume of a spherical wedge is equal to the lune which forms its base, multiplied by one-third of the radius.

PROPOSITION XVI. THEOREM.

Symmetrical triangles are equal in area.

Let ABC and DEF be symmetrical triangles, the side DE being equal to AB, the side DF to AC, and the side EF to BC: then will the triangles be equal in

For, conceive a small circle to be drawn through A, B,

and C, and let P be its pole; draw arcs of great circles from P to A, B, and C: these

:

equal to AC, by hypothesis; the side FQ equal to PC by construction, and the angle DFQ equal to ACP, by construction hence (P. VIII.), the side DQ is equal to AP, the angle FDQ to PAC, and the angle FQD to APC. Now, because the triangles QFD and PAC are isosceles and equal in all their parts, they may be placed so as to coincide throughout, the base FD falling on AC, DQ on CP, and FQ on AP: hence, they are equal in area.

If we take from the angle DFE the angle DFQ, and from the angle ACB the angle ACP, the remaining angles QFE and PCB, will be equal. In the triangles FQE and PCB, we have the side QF equal to PC, by construction, the side FE equal to BC, by hypothesis, and the angle QFE equal to PCB, from what has just been shown: hence, the triangles are equal in all their parts, and being isosceles, they may be placed coincide throughout, the side QE falling on PC, and the side QF on PB; these triangles are, therefore, equal in

area.

as to

In the triangles QDE and PAB, we have the sides QD, QE, PA, and PB, all equal, and the angle DQE equal to APB, because they are the sums of equal angles: hence, the triangles are equal in all their parts, and

because they are isosceles, they may be so placed as to coincide throughout, the side QD falling on PB, and the side QE on PA; these triangles are, therefore, equal in

area.

Hence, the sum of the triangles QFD and QFE, is equal to the sum of the triangles PAC and PBC. If from the former sum we take away the triangle QDE, there will remain the triangle DFE; and if from the latter sum we take away the triangle PAB, there will remain the triangle ABC: hence, the triangles ABC and DEF are equal in area; which was to be proved.

Scholium. If the point P falls within the triangle ABC, the point will fall within the triangle DEF. In this case, the triangle DEF is equal to the sum of the triangles QFD, QFE, and QDE, and the triangle ABC is equal to the sum of the equal triangles PAC, PBC, and PAB; the proposition, therefore, still holds good.

PROPOSITION XVII. THEOREM.

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

Let the circumferences AOB, COD, intersect on the surface of a hemisphere then will the sum of the opposite triangles AOC, BOD, be equal to the lune whose angle is BOD.

For, produce the arcs OB, OD, on the other hemisphere, till they meet at N. Now, since AOB and OBN

A

are semi-circumferences, if we take away the common part

OB, we shall have BN equal to 40.
son, we have DN equal to CO, and BD
hence, the two triangles AOC, BDN,
have their sides respectively equal:
they are therefore symmetrical; con-
sequently, they are equal in area
(P. XVI.).
But the sum of the tri-
angies BDN, BOD, is equal to
the lune OBNDO, whose angle is
BOD: hence, the sum of AOC and
BOD is equal to the lune whose
angle is BOD; which was to be proved.

A

For a like rea

equal to AC:

Schoum. It is evident that the two spherical pyramids, which have the triangles AOC, BOD, for bases, are together equal to the spherical wedge whose angle is BOD.

PROPOSITION XVIII. THEOREM.

The area of a spherical triangle is equal to its spherical excess multiplied by a tri-rectangular triangle.

Let ABC be a spherical triangle: then will its surface 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, AGII, are together equal to the lune whose angle is A; but the area of this lune is equal to 24 x T (P. XV., C. 2):

hence, the sum of the

triangles ADE and AGH, is equal

to 24 × T. In like manner, it may be shown that the