(E) COROLLARY. If a side and the two adjacent angles ̧ in one spherical triangle, be equal to a side and the two adjacent angles in another, each to each; these triangles are equal in all respects.

For by taking the supplements of each of the two angles, and of the sides contained between them, two new triangles will be formed which will be equal, by the proposition, and therefore their supplements, viz. the original triangles will be equal.

PROPOSITION X.

(F) The angles at the base of an isosceles spherical triangle are equal to each other.

DEMONSTRATION. Let ABC be an isosceles triangle having the side Ac equal to the side BC.

Make CE equal to CD, and draw EB and DA; now AC being equal to BC by hypothesis, and Ec equal to CD by construction, and the angle ACB common to

E

B

both the triangles ACD and BCE, these two triangles are equal in all respects (D. 141.); therefore AD is equal to EB.

Again, since ac is equal to BC, and Ec equal to CD, the remaining parts AE and DB are equal, hence the triangles BAE and ABD are equal in all respects; for the two sides AE and EB are equal to the two sides BD and DA each to each, and the base AB is common to both, therefore (B. 140.) the angle BAE is equal to the angle ABD. Q. E. D.

(G) COROLLARY I. In any spherical triangle if two of the angles be equal, the sides opposite to them will be equal.

Make BD equal to AE, and draw EB and AD, then the triangle BAE is equal to the triangle ABD (D. 141.); therefore AD is equal to BE, and the angle AEB equal to the angle ADB, the angle DAB equal to the angle EBA; now the angle BEC is equal to the angle ADC, being each of them supplements of the equal angles AEB and BDA, therefore the triangle ADC is equal to the triangle BEC (E. 142.); consequently AE+EC is equal to ED+DC. viz. equal sides are opposite to equal angles.

(H) COROLLARY II. Hence every equilateral triangle is likewise equiangular, and the contrary.

(I) COROLLARY III. A line drawn from the vertex of an isosceles triangle to the middle of the base, is perpendicular to the base.

For the two sides FB and BC are equal to the two sides FA and AC, and the angle FBC is equal to the angle FAC, therefore the angle CFB is equal to the angle CFA (D. 141.); but CFA and CFB

are equal to two right angles (M. 134.), therefore cr is perpendicular to AB.

(K) COROLLARY IV. A perpendicular drawn from the vertex of an isosceles triangle upon the base, always bisects the base, and the vertical angle, except when the two sides are quadrants, in which case there may be an indefinite number of perpendiculars.

For the triangles FBC and FAC are equal in all respects. (D. 141.) But if AC and BC are quadrants, that is, if c is the pole of the base, then an infinite number of lines may be drawn from the point cupon the base to cut it at right angles. (H. and I.134.)

PROPOSITION XI.

(L) In any spherical triangle (ABC) the greatest side (AC) is opposite to the greatest angle (B), and the least to the least.

DEMONSTRATION. Make the angle DBA equal to the angle BAD, then AD is equal to BD. (G. 142.) Now AD+DC is A equal to AC, but AD + DC or BD + DC is

B

D

greater than BC (Y. 138.), therefore ac is greater than BC. Lastly, DC is less than AD or AB, and it may be shewn, as above, that the angle DBC is less than the angle DCB.

PROPOSITION XII.

Q. E. D.

(M) If one side (DC) of a spherical triangle (DCB) be produced, and if the sum of the other sides (CB+BD) be greater, equal to, or less than a semi-circle; the internal angle (D) at the base (DC) is accordingly greater, equal to, or less than the outward and site angle (BCA) adjacent to the side produced.

DEMONSTRATION.

Produce the

sides DB and DC till they meet in a. Then the arcs ABD and ACD are each of them a semi-circle (C. 133.), and the angle at D is equal to the angle at A. (Q. 135.)

B

oppo

D

1. If CB+BD be greater than 180°, BC will be greater than AB, and the angle CAB, or, which is the same thing, the angle D will be greater than the angle BCA. (L. 143.)

2. If CB+BD be equal to 180°, then BA and BC will be equal, and the angle CAB will be equal to the angle BCA, or the angle D will be equal to the angle BCA. (F. 142.)

3. If CB+BD be less than 180°, BC will be less than AB, and the angle CAB, or the angle D will be less than the angle ACB. (L. 143.) 2. E. D.

(N) COROLLARY. In any spherical triangle, if the sum of any two sides be greater, equal to, or less than a semi-circle; the sum of the opposite angles will be greater, equal to, or less than a semi-circle.

For, according as CB+BD is greater, equal to, or less than a semi-circle, the angle D opposite to the side BC is greater, equal to, or less than the angle BCA; but the angle BCA and the angle BCD, opposite to the side DB, are always equal to two rightangles (M. 134.); therefore whenever D is greater than BCA, the sum of the angles BCD and BDC is greater than two rightangles; et contra.

PROPOSITION XIII.

(O) A right-angled spherical triangle may have either, 1. One right angle, and two acute angles.

2. One right angle, and two obtuse angles. 3. One obtuse angle, and two right angles.

4. One acute angle, and two right angles. OR, 5. Three right angles.

DEMONSTRATION. Let the angles A and D be right angles; and the arcs AE and ED quadrants; then E will be the pole of the circle ACGD, and Ecand EG will be quadrants (H. 133.) Since AE and DE are quadrants, a or D will be the pole of EG; therefore AG and GD are quadrants.

A

B

E

H

D

1. Because AE and Ec are quadrants, the angles EAC and ACE are right angles (I. 134.); therefore the angle ACB is less than a right angle. And since AC is less than AG, the angle ABC is less than the angle AEG (L. 143.); but AEG is a right-angle; therefore the right-angled triangle CAB has two acute angles and one right-angle.

2. The triangle BDC right-angled at D, has two obtuse angles and one right angle. For the angles DBC and DCB are supplements of the two acute angles ABC and ACB, and therefore must be obtuse.

3. The triangle CED has one obtuse angle, and two right angles. For, since GD and ED are quadrants, the angles DEG and DGE are right angles, therefore DEC is an obtuse angle. But Ec is a quadrant, and E is the pole of CGD, therefore ECD is a right angle. (H. 133.)

4. The triangle AEC has already been shewn to have two right angles; and AEG is a right angle, but AEC is less than a right angle.

5. The right-angled spherical triangle GAE has three right angles, for a is a right angle by the proposition, and a is the pole of the circle EG, therefore AEG and AGE are right angles. (H. 135.) Q. E. d.

(P) COROLLARY I. In any right-angled triangle, if the two legs be of the same species, viz. both acute, or both obtuse, the hypothenuse will be less than a quadrant, and the contrary.

For BC the hypothenuse, or side opposite to the right angles A and D, is common to the triangle BAC having two acute angles, and to the triangle BDC having two obtuse angles.

(Q) COROLLARY II. In any right-angled spherical triangle, if the legs be of different species, the hypothenuse will be greater than a quadrant, and the contrary: but one of the legs, at least, will be a quadrant, if the hypothenuse be a quadrant.

For, in the triangle ICD, right-angled at c, ci is less than a quadrant and CD greater, from what has already been demonstrated; but since D is the pole of the circle EHG, DH is a quadrant, therefore DI the hypothenuse is greater than a quadrant. And when ci becomes equal to CE, DI will become equal to DE; but DE is a quadrant, therefore if the hypothenuse be a quadrant, one of the legs, at least, will be a quadrant. But both the legs will be quadrants when all the angles are right angles, as in the triangle AEG.

(R) COROLLARY III. The sides containing the right angle of any triangle, are always of the same species as their opposite angles, and the angles are of the same species as their opposite sides.

For in the triangle CAB where the two angles are acute, the opposite sides AB and AC are acute; and in the triangle CDB where the two angles are obtuse, the opposite sides BD and CD are obtuse.

(S) COROLLARY IV. If the hypothenuse be less than 90°, the legs are of the same species as their adjacent angles. If the hypothenuse be greater than 90°, the legs and their adjacent angles are of different species.

This is evident from the triangles BAC, CDB, and Dci.

(T) COROLLARY V. According as the hypothenuse, and one side (or its opposite angle) are of the same, or different species; the other side (and its opposite angle) will be less or greater than a quadrant.

This is also evident, from the triangle Dci.

PROPOSITION XIV.

(U) If any number of arcs of circles CPB, CmE, CNF, CDA, CSH, COG, be drawn from one and the same point c, on the surface of the sphere, to the circumference of a great circle AHGFEB; the greatest arc will be that which passes through the pole P.

L

of the great circle AHGFEB; and the least arc will be the supplement ADC of the greatest CPB. And that arc which is the nearest to the greatest will be greater than any one more remote; likewise, that arc which is nearest to the least arc, will be less than any one more remote from it.

DEMONSTRATION. Let z be the centre of the sphere, or of the great circle AHGFEB, and AB its diameter; from c, the point where all the circles intersect each other, draw ci perpendicular to AB, and join IE, IF, IG and Iн.

Then of all the straight lines which can be drawn from the point I to the circumference of the circle AHGFEB, that line IB, is the greatest which passes through the centre z, and its opposite line IA is the least; that line IE, which is nearest to IB, is greater than the line IF which is more remote; and the line IH which is nearest to IA, is shorter than the line IG which is more remote. (Euclid 7 of III.)

Let AC, HC, GC, FC, &c. be joined by straight lines; then will CIB, CIE, CIF, CIG, CIH and CIA be all right-angled plane triangles, having the perpendicular ci common to each; therefore that triangle CIB, which has the greatest base IB, will have the greatest hypothenuse BC; and that tri

A

H

m

angle CIA which has the least base IA, will have the least hypothenuse AC, &c.; of the others, EC will be greater than Fc, and HC less than GC.

But these several hypothenuses are all chords of equal circles, and the greater chord cuts off the greater arc, or circumference, therefore CPB is the greatest arc, and ADC the least; cmx greater than CnF, and CSH is less than COG.

is

PROPOSITION XV.

C

Q. E. D.

(W) If the angles at the base of an oblique-angled spherical triangle be of the same species, viz. both acute, or both obtuse, a perpendicular from the vertical angle upon the base will fall within the triangle; if they be of different species, it will fall without. DEMONSTRATION. In the rightangled triangle CDB, CD is of the same species as CBD (R.145.), and in the right-angled triangle CDA, CD is of the same species as CAD, therefore the perpendicular in the A first figure falls between the angles A and B.

I.

B

II.