shall be equal, each to each, viz. those to which the equal angles are opposite, and the third angle of the one shall be equal to the third angle of the other.

The case in which the equal angles lie in the same direction from one another in the two triangles may be demonstrated by superposition, in the same manner as Book I. Prop. 5, to which the reader is referred. The other case, in which they lie in different directions, may be demonstrated as follows:

Let A B C, DEF (see the figure of Prop. 13.) be two spherical triangles, which have the two angles AB C, ACB of the one equal to the two angles DEF, DFE of the other, each to each, but in a reverse order, and likewise the side B C equal to the side EF: their other sides shall be equal each to each, and the third angle B A C shall be equal to the third angle EDF.

Let there be described, as in the last proposition, upon the other side of the base BC the triangle A' B C, which has its sides and angles equal to those of the triangle ABC, each to each, but in a reverse order. Then, because the angles A'B C and D E F are equal, each of them, to the angle A B C, they are equal to one another; and, for the like reason the angle A'CB is equal to the angle DFE. Also, BC is equal to EF; and these equal parts lie in the same direction from one another in the two triangles A' B C, DEF. Therefore, by the first case, the angle B A' C is equal to the angle E D F, and the sides A' B, A'C are equal to the sides D E, D F respectively. And, because the angle BAC is equal to the angle B A'C, and the sides AB, AC to the sides A'B, A' C, each to each, the angle B A C is equal to the angle EDF, and the sides AB, AC to the sides DE,DF respectively. Therefore, &c.

PROP. 15.

If two spherical triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, the angle contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them of the other.

The case in which the equal sides lie in the same direction from one another, may be demonstrated in the same manner as Book I. Prop. 7: the other case, in which they lie in different directions,

as follows:

Let A B C, DEF (see the figure of

Prop. 13) be two spherical triangles having the two sides of the one equal to the two sides of the other, each to each, and likewise the base B C equal to the base EF. The angle BAC shall be equal to the angle EDF.

Let there be described, as in Prop. 13, upon the other side of the base B C the triangle A' B C, which has its sides and angles equal to those of the triangle ABC, each to each, but in a reverse order. Then, because A' B and D E are each of them equal to AB, they are equal to one another: and, for the like reason, A'C is equal to DF: also B C is equal to E F; and these equal parts lie in the same direction from one another in the two triangles A' B C, DE F. Therefore, by the first case, the angle BA'C. is equal to the angle ED F. Therefore, because the angle B A C is equal to the angle B A' C, the angle BAC is likewise equal to the angle EDF.

the three angles A', B', C' of the one equal to the three angles D', E', F' of the other, each to each. And hence the triangles A B C, DEF have the three sides of the one equal to the three sides of the other, each to each, because every two corresponding sides, as AB and DE are supplements of the measures of equal angles C' and F' (7.) Therefore, &c.

PROP. 17.

If two spherical triangles have two sides of the one equal to two sides of the other, each to each, but the angle which is contained by the two sides of the one greater than the angle which is contained by the two sides which are equal to them of the other; the base of that which has the greater angle shall be greater than the base of the other: and conversely. See the demonstration of Book I. Prop. 11.

PROP. 18.

If any point be taken within a circle of the sphere which is not its pole; of all the arcs which can be drawn from that point to the circumference, the greatest is that in which the pole is, and the other part of that arc produced is the least; and of any others, that which is nearer to the greatest is always greater than one more remote: and from the same point to the circumference there can be drawn only two arcs that are equal to one another, which two make equal angles, upon either side, with the shortest arc.

A

G

E

D

F

C

Let A B C be any circle (either a great or a small circle) of the sphere, P its pole (if a great circle, either of its two poles; if a small circle, the nearer pole), and D a point on the surface of the sphere, which is upon the same side of the circle A B C D with the pole P. Of all the arcs D A, D P B, D C, &c. which can be drawn from D to the circumference, the arc D PB which passes through P ́shall be the greatest, and DA the other part of that are produced the least: and of the others, DC, which is nearer to D P B, shall be greater than DE, which is more remote.

Join PE, PC. Then, because the two sides D P, P C of the triangle D PC are together greater than the third DC (9.), and that PB is equal to PC, the arc

DPB is greater than D C. Again, because the triangles PCD, PED have the two sides PC, PD of the one equal to the two sides PE, PD of the other, each to each, but the angle CPD greater than the angle EPD, the base DC is greater than the base DE (17.). Again, because the side ED of the triangle PDE is greater than the difference of PD, PE the other two sides (9.), and that PA is equal to PE, the arc ED is greater than the difference of PD and PA, that is, than AD. Therefore, DPB

is the greatest and D A the least of all circumference, and D C, which is nearer to DPB, is greater than DE, which is

arcs which can be drawn from D to the

more remote.

equal arcs DE, DF from the point D Also, there can be drawn only two

to the circumference, and these make equal angles with the shortest arc D A upon either side of it. For, if PF be drawn, making the angle APF equal to the angle APE, and D F be joined; then, because the triangles PDF, PDE have two sides of the one equal to two sides of the other, each to each, and the included angles DP F, DPE equal to one another, the bases D F and DE are equal to one another, and the angles PDF, PDE which are adjacent to ADF, ADE, likewise equal (13.). But, besides D F, there cannot be drawn any other arc from D to the circumference equal to DE; for if it were possible, and D G were such an arc, then DG, which is nearer to (or farther from) DPB than D F is, would be equal toD F, which is contrary to what was shown in the former part of the proposition.

Therefore, &c.

Cor. 1. From a point to a spherical arc, the perpendicular is either the least or the greatest distance, the least when it is less than a quadrant, the greatest when it is greater than a quadrant. For the perpendicular arc either passes, or may be produced to pass through the pole of the spherical arc (5. Cor. 1.): in the former case it is, as DPB, greater than any other arc which can be drawn from the point E to the circumference of which the spherical arc is a portion, and it is, at the same time, greater than (5.) the quadrant PB ; in the latter case it is, as DA, less than any other, and it is, at the same time, less than (5.) the quadrant P A.

Cor. 2. If there be taken within a circle (that is, anywhere on the surface of the sphere, if it be a great circle, or

on the same side with the nearer pole, if a small circle) a point from which there fall more than two equal arcs to the circumference, that point shall be a pole of the circle.

Cor. 3. Hence, also, if there be taken without a small circle a point from which there fall more than two equal arcs to the circumference, that point shall be the more distant pole of the small circle. For if the opposite point through which the equal arcs pass (4 Cor. 2.) be taken, three distances of the latter from the circumference (I. ax. 3.) will be equal, and therefore, by Cor. 2., it will be the nearer pole.

PROP. 19.

In a spherical triangle, which has one of its angles a right angle, either of the other angles is greater or less than a right angle, according as the opposite side is greater or less than a quadrant; and if the sides, which contain the right angle, are both greater or both less than quadrants, the third side, or hypotenuse, shall be less than a quadrant; but if one of these sides be greater than a quadrant, and the other less, the hypotenuse shall be greater than a quad

rant.

For, if from BA, or from BA produced, BP be cut off equal to a quadrant, and PC joined, the angle P C B will be a right angle, because Pis the pole of B C (5.); and it is evident that the angle A CB is greater or less than the angle PC B, according as AB is greater or less than PB.

Also, if the sides A B, BC are both greater or both less than quadrants, the hypotenuse AC shall be less than a quadrant; but if one of them be greater than a quadrant, and the other less, A C shall be greater than a quadrant.

For, if from B C, or from BC produced, BD be cut off equal to a quadrant, and AD be joined, D will be the pole of A B (5.), and, therefore, A D will be equal to a quadrant. And, first, if AB and B C be both less than quadrants,

then, A B being (18. Cor. 1.) the least arc which can be drawn from A to B C, A C which is nearer to it is less than AD, which is more remote (18.); or, secondly, if AB and BC be both greater than quadrants, then A B being (18. Cor. 1.) the greatest arc which can be drawn from

B

B

A to BC, AC which is further from it is less than AD,which is not so remote (18.); therefore, in both cases, A C is less than a quadrant. But, if either of the sides, as AB, be greater than a quadrant, and the other, BC, less, AB being the greatest are which can be drawn from A to BC, A C which is nearer to it is greater than AD, which is more remote 18.); therefore, A C is greater than a quadrant.

Therefore, &c.

Cor. 1. It is evident from the demon

stration, that if one of the two sides which contain the right angle be equal to a quadrant, the opposite angle will be a right angle, and the hypotenuse likewise a quadrant.

Cor. 2. Hence, in a right-angled spherical triangle, either of the two sides is of the same affection with the opposite angle; and the hypotenuse is, 1o, equal to, or, 20, less than, or, 3°, greater than, a quadrant, according as, 1°, one of the adjoining angles is a right angle, or these two angles are, 20, of the same or, 3°, of different affections. See note, page 61.

Spherical triangles which have the three sides of the one equal to the three sides of the other, each to each, contain also equal portions of spherical surface, whether the triangles are symmetrical or otherwise.

Let A B C, DEF be two spherical triangles, which have the three sides of the one equal to the three sides of the other, each to each, viz. A B to DE, AC to D F, and B C to EF: the surface of the triangle ABC shall be equal to the surface of the triangle D E F.

Let a plane be drawn through the points A, B, C, to cut the sphere in a circle passing through those points (1.): from the centre of the sphere, draw a perpendicular to the plane ABC, and produce it to meet the surface of the sphere in P: then P is the pole of the circle A B C (def. 3.); and, consequently, (2.) if the spherical arcs PA, PB, PC are drawn, joining the point P with the points A, B, C, respectively, they will be equal to one another. Make the spherical angle DEQ equal to the spherical angle ABP: make also the arc EQ equal to the arc B P, and join QD, Q F. Then, because the angles DEF (15.) and D E Q are equal to the angles A B C and A B P, each to each, the angle QEF is equal to the angle PBC (I. ax. 3.). And, because in the triangles Q E F, PB C, two sides of the one are equal to two sides of the other, each to each, and the included angles QEF, PBC likewise equal to one another (13.) the arc QF is equal to the arc PC, and the angle EQ F to the angle B PC. In the same manner, it may be shown that the arc Q D is equal to the arc PA, and the angle DQE to the angle AP B. Therefore, because the angles DQE and EQF are respectively equal to the angles APB and BP C, the angle DQF is equal to the angle APC: and because the arcs PA, PB, PC are equal to one another, the arcs Q D, QE, Q F, which are severally equal to them, are likewise equal to one another (I. ax. 1.).

Now, it is evident that isosceles triangles, as P B C and QEF, which have equal vertical angles, and equal sides containing them, may be made to coincide, and, therefore, are equal to one another. Therefore, the isosceles triangles PAB, PB C, and PAC are respectively equal to the isosceles triangles QDE, QEF, and QD F. But the triangle ABC is equal to the sum of

the former three, if the point P falls within the triangle A B C ; or to the difference between the sum of two of them and the third, if it fall without the triangle; and the triangle D E F is equal to the sum of the latter three, in the first case, or to the difference between the sum of the corresponding two of them and the third, in the other*. Therefore, the triangle ABC is equal to the triangle DEF (I. ax. 2. 3.). Therefore, &c.

surface, and the sum of three lunes having their angles respectively equal to the angles A, B, C.

Let the circles be completed, of which the sides A B, A C, and B C are parts, and let A a, Bb, and C c be the diameters passing through the points A, B, and C. Then, because C Ac and A ca are semicircumferences (4.), they are equal to one another, and A c being taken from each, the remainders AC, ac are likewise equal. In the same manner it may be shown that A B is equal to a b, and B C to bc. And, because the triangles ABC, abc have the three sides of the one equal to the three sides of the other, each to each, their surfaces are equal to one another (20.) Now, the lune A cab A has its angle equal to the angle A of the triangle A B C (8. Cor. 1.), and the lunes B Ab C B, CB CAC have for their angles the angles B, C of the same triangle. And these three lunes are

*If the arcs B P, E Q, or these arcs produced, cut the arcs AC, DF in the points R, S, respectively, it may be shewn by prop. 14, that BR is equal to ES: therefore, since BP is equal to EQ, if the point P falls within the triangle ABC, the point Q must likewise fall within the triangle DEF; or if P fall

without the triangle ABC, and within the angle ABC, Q must likewise fall without the triangle

DEF, and within the corresponding angle DE F.

together equal to the surface of the hemisphere A B C bc together with the triangles ABC, abc; that is, to the surface of the hemisphere, together with twice the triangle ABC. Therefore, twice the triangle A B C is equal to the difference between the sum of the three lunes and the surface of the hemisphere; and the triangle ABC is equal to half that difference.

Therefore, &c.

Cor. 1. Hence, if the angles of a spherical triangle are given, we may find the proportion which its surface bears to the surface of the sphere. For, since the surface of any lune is to the surface of the sphere, as the angle of the lune to four right angles (V. 21.), or as twice its angle to eight right angles; if the surface of the sphere be represented by 8 right angles, or 8 R,* the surfaces of the lunes which have the angles A, B, and C will be represented by 2 A, 2 B, and 2 C; and the hemispherical surface will be represented by 4 R: therefore, the surface of the triangle A B C will be represented by half 2 A+ 2 B+2 C-4 R, that is, by A+B+C2 R. And hence, it is commonly said, that the surface of a spherical triangle is measured by the excess of the sum of its angles above two right angles; it being understood that the surface of the sphere is measured by eight right angles.

Cor. 2. The surface of a spherical triangle whose angles are A, B, C is equal to the surface of a lune whose A+B+C angle is

2

R.

For both surfaces are measured by A + B + C

2 R.

For the polygon may be divided into as many triangles as it has sides, by joining its angles with any point P taken within the polygon; and the surface of each of these triangles is measured by

* In which case, R represents the surface of the spherical triangle, which has each of its angular points the pole of the side opposite to it (7. Scholium.). We shall see (28. Scholium.) that the surface of this triangle measures a solid right angle at the centre of the sphere.

Scholium.

It follows from the last corollary, that if S be the number of the solid angles of a polyhedron, F the number of its faces, and E the number of its edges, S+F-E will be equal to 2. For, if a sphere be described about any point within the polyhedron as a centre, the lines which are drawn from this point through the solid angles of the polyhedron will cut the surface of the sphere in S points, and the planes which are drawn from the same point through the edges will cut the surface in E spherical arcs; and thus the whole surface of the sphere will be divided into F spherical polygons, corresponding to the F faces of the polyhedron. Now, the surface of each of these spherical polygons is measured by the sum of its angles plus four right angles minus twice as many right angles as the polygon has sides. Therefore, the surfaces of all the polygons taken together are measured by the sum of all their angles taken together plus as many times four right angles as there are polygons minus twice as many right angles as all the polygons together have sides. But the sum of all their angles taken together is the sum of all the angles, at the S angular points, and, therefore, is equal to S times four right angles (8. Cor. 2.): the number of the polygons is F; and the number of the sides of the polygons is 2 E, because the number of different sides is E, and each of these is at once a side of two adjoining polygons. Therefore, the surfaces of all the polygons, taken together, are measured by (S+FE) times four right angles. Therefore, since the surface of the