And, because BM and CN are quadrants, these arches together, or MN and CB together, are equal to a semicircle; and therefore MN, the measure of the angle FDE, is the complement of the opposite side CB of the triangle ACB, to a semicircle. In like manner it may be demonstrated, that KL the measure of the angle DEF, is the complement of the side AC, and GH the measure of the angle F, of the side AB, to a semicircle.

Scholium. Although, in this proposition, the word complement is used in its customary meaning; yet the complement of an arch or angle, to a semicircle or two right angles, is also called its supplement, as mentioned in Def. 5. Pl. Tr.; and therefore one of the triangles mentioned in this proposition, is said to be supplemental of the other.

PROP. XIV. THEOR.

The three angles of a spherical triangle (ABC, see fig. to the preceding proposition), are greater than two right angles, and less than

six.

For the three measures of the angles of the triangle ABC, together with the three sides of its supplemental triangle DEF, are equal to three semicircles (13. Sph. Tr.); but the three sides of the triangle DEF are less than two semicircles (11. Sph. Tr.); therefore the three measures of the angles of the triangle ABC, are greater than a semicircle, and of course these three angles are together greater than two right angles.

And these three angles are less than six right angles, for the internal and external angles together of the triangle ABC, are equal to six right angles (4. Sph. Tr.), therefore the three interkal angles are less than six right angles.

PROP. XV. THEOR.

Of all arches of great circles (DA, DE, DF and DB), which can be drawn to any great circle (AEB) of a sphere, from any point (D) which is not its pole, the greatest is that (DA) which passes through the pole (H), and the continuation of it (DB) is the least; and of others (DE and DF), that (DE) which is nearer to the greatest, is greater than the more remote (DF); and the angles ( DEA and DEB) which any of them (as DE), which does not pass through its pole H), makes with it, are unequal, that (DEA) which is towards the pole, being the greater.

Let AB be the common section of the circles ADB and AEB, from D draw DG perpendicular to AB, and join GE, GF, DA, DE, DF and DB.

Because AB is a diameter of the sphere (1. Sph. Tr.), its mid- A dle point is the centre of the same, and therefore a right line drawn thereto, from the pole H of the circle AEB, is perpendi

H

D

B

E

K

cular to the plain AEB (Cor. 2. 2 Sph. Tr.), and therefore to AB (Def. 3. 2 Sup.), and therefore, DG being perpendicular to AB (Constr.), parallel to DG (28. 1 Eu.); whence, the right line so drawn from H to the centre being perpendicnlar to the plain AEB, the right line DG is perpendicular to the same plain (5. 2 Sup.),, and so the angles DGA, DGE, DGF and DGB are right angles (Def. 3. 2 Sup).

And GA is the greaiest, and GB the least, of all right lines, which can be drawn from G to the circumference AEB, and the right line GE greater than GF (7. 3 Eu.); whence, the triangles DGA, DGE, DGF and DGB being right angled at G, and having the side DG common, the square of DG and GA together, or, which is equal (47. 1 Eu.), the square of DA, is greater than the squares of DG and GE together, or which is equal (by the same), the souare of DE; therefore the right line DA is greater than DE. In like manner it may be shewn, that the right line DE is greater than DF, and DF than DB; whence it follows, that the arch DHA is the greatest, and DB the least, of all arches of great circles, which can be drawn from D to the great circie AEB, and the arch DE, which is nearer to DHA, greater than the arch DF which is more remote.

And an arch of a great circle being supposed to be drawn from H to E, the spherical angle HEA is a right angle (Cor. 7. 2 Sph. Tr.), therefore the spherical angle DEA is greater, and of course (4. Sph. Tr.), DEB less than a right angle. In like manner it may be proved, that the spherical angle DFA is greater, and DFB less than a right angle.

Cor. 1. Hence it follows, that to any great circle [AEB] of a sphere, from any point [D] which is not its pole, but two perpendiculars [DHA and DB] can be drawn, one [DHA] passing through its pole [H], and the other [DB] the continuation thereof, their incidences being at the interval of a semicircle from each other.

Cor. 2. If to a great circle [AEB] of a sphere, from any point [D] without it which is not its pole, an arch [DK] of a great circle be drawn, to the point [K] in which an arch [AEB], intercepted between the perpendiculars [DHA and DB] let fall from that point on the great circle, is bisected; the arch [DK] so drawn is a quadrant.

For the angle KBD being a right angle, and KB a quadrant, K is the pole of the circle BD (Cor. 6. 2 and prop. 2 Sph. Tr.), and therefore DK is a quadrant (2. Sph. Tr).

PROP. XVI. THEOR.

The legs containing the right angle of a right angled spherical triangle, are of the same affectiou, as the opposite angles. That is, according to the legs are equal to, or greater or less than, quadrants, the angles opposite to them are equal to, or greater or less than, right angles.

Part 1. Let the spherical triangle HAK (see fig. to the prec, prop.), right angled at A, have the leg AH equal to a quadrant, K being supposed to be any where in the circle AEB, H is the pole of the arch AK (Cor. 6. 2 and prop. 2 Sph. Tr.), therefore the angle HKA is a right angle (Cor. 7.2 Sph. Tr).

Part 2. Let the spherical triangle DEA or DFA, right angled at A, have a leg, as AHD, greater than a quadrant, and the angle DEA or DFA, opposite that leg, is obtuse (15. Sph. Tr). Part S.

Let the spherical triangle DFB or DEB, right angled at B, have a leg, as BD. less than a quadrant, andthe angle DFB or DEB, opposite that leg, is acute (15 Sph. Tr).

PROP. XVII. THEOR.

If two legs of a right angled spherical triangle be of the same affection, (and consequently by the preceding proposition, the angles opposite to them); the hypothenuse is less than a quadrant; unless both the legs be quadrants, in which case the hypothenuse is a quadrant; if they be of different affections, the hypothenuse is greater than a quadrant.

Part 1. If in the spherical triangle ADF, see fig. to proposi tion 15, right angled at A, the legs AD and AF be both greater than quadrants, or in the triangle BDF, right angled at B, the legs BD and BF be both less than quadrants; the hypothenuse DF is less than a quadrant.

For the semicircle AEB being bisected in K, the arch of a great circle DK being drawn is a quadrant (Cor. 2. 15 Sph. Tr.), and DF is less than DK (15. Sph. Tr.), therefore DF is less than a quadrant.

Part 2. If in the spherical triangle AHK, right angled at A, the legs AH and AK be quadrants; the third side HK is a quadrant.

For A is the pole of the circle HK (Cor. 3. Sph. Tr.); whence, the spherical angle KAH being a right angle (Hyp.), the arch HK is a quadrant (3 Sph. Tr).

Part 3. If in the spherical triangle ADE, right angled at A, one of the legs AD be greater, and the other AE less than a quadrant; DE is greater than a quadrant.

For the arch DK being drawn as in part 1, is a quadrant; and DE is greater than DK (15 Sph. Tr.); therefore DE is greater than a quadrant.

PROP. XVIII. THEOR.

According as the hypothenuse of a right angled spherical triangle is greater or less than a quadrant; the legs, (and consequently by 16. Sph. Tr. the oblique angles), are of different affections, or of the same.

Part 1. If the hypothenuse be greater than a quadrant, the legs are of different affections.

For if the legs were of the same affection, the hypothenuse would not be greater than a quadrant (17. Sph. Tr.), contrary to the supposition.

Part 2. If the bypothenuse be less than a quadrant, the legs are of the same affection.

For if they were of different affections, the hypothenuse would be greater than a quadrant (17 Sph. Tr.), contrary to the supposition.

PROP. XIX. THEOR.

If, in any spherical triangle, (ABC, see figure 1 and 2), the perpendicular (CD) let fall on the base (AB) from the opposite angle, fall within the triangle (as in figure 1); the angles at the base are of the same affection: if the perpendicular fall without the triangle (as in fig. 2); the angles at the base are of different affections.

Part 1. Fig. 1. Since the triangles ACD and BCD are both right angled at D, the angles A and B are of the same affection with CD (16 Sph. Tr.), and therefore with each other.

Part 2. When the perpendicular CD falls without the triangle, as in fig. 2, the angles A and GBD are of the same affectiou with CD (16 Sph. Tr.), and therefore with each other, and the angles CBA, and CBD are of different affections (4. Sph. Tr.) ; therefore the angles A and ABC are of different affections.

Scholium. In part 2, it is supposed, that CD is not equal to a quadrant, for if it were, C would be the pole of the arch ABD (Cor. 6. 2 and prop. 2 Sph. Tr.), and all the angles CAB, CBA and C3D would be right angles (Cor. 7 2 Sph. Tr). A like observation is applicable to the next proposition, the case being excepted, when the vertex of the triangle is the pole of the base