PROP. XVII.

(111.) Theorem. If two spherical triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by those two sides of the one, greater than the angle contained by the two sides equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other.

The proposition is proved, by the help of the preceding articles, in the same manner as is the twentyfourth proposition of the first book of Euclid's Elements.

(112.) COR. From the proposition itself may be deduced its converse; in the same manner as the twentyfifth proposition of the first book of Euclid's Elements is deduced from the twenty-fourth of that book.

PROP. XVIII.

(113.) Theorem. If two spherical triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, the sides opposite to equal angles in each, and if the sides subtending the other two equal angles be, together, unequal to two quadrants, then, the remaining sides shall be equal, each to each; and the third angle of the one, to the third angle of the other.

Let ABC, EFG be two spherical, triangles, having

the angle A equal to E, the angle C equal to G, and the sides BC, FG, which subtend equal angles, also equal :

and let the two sides BA and FE, which subtend the other equal angles, be, together, either greater, or less, than two quadrants: then shall BA be equal to FE, AC to EG and the angle ABC to the angle EFG.

For, if AC be unequal to EG, let AC be the greater; from AC cut off CD equal to EG, and join (Art. 66.) B, D; then, since the two sides BC, CD are severally equal to FG, GE, and the angle BCD to FGE, therefore (Art. 97.) the angle BDC is equal to FEG, and BD to FE; but the angle FEG, by the supposition, is equal to BAC; wherefore, the angle BDC is equal to BAC; and consequently, (Art. 100.) AB and BD are, together, equal to a semi-circumference; but BD has been proved to be equal to FE: wherefore, BA and FE are, together, equal to a semi-circumference: which is contrary to the hypothesis. So that AC and EG are not unequal; therefore, (Art. 97.) BA is equal to FE, and the angle ABC to EFG.

PROP. XIX.

(114.) Theorem. If two spherical triangles have one angle of the one equal to one angle of the other, and have also the two sides about another angle in each severally equal, and if the third angle, in each, be either greater or less than a right angle, the remaining side of the one triangle shall, also, be equal to the remaining side of the other, and the other angles to the other angles, each to each.

This may be inferred from Art. 113. by the help of polar triangles or it may be proved, ex absurdo, the same construction having first been made, as in Art. 113.

:

For, let the angle C in the triangle ABC, be equal to G, in EFG; also let AB, BC be equal to EF, FG,

B

F

A

D

each to each; and let the angles A and E not be right angles: then if AC be not equal to EG, make CD equal to it, and join B, D. Then (Art. 97.) BD is equal to FE, and the angle BDC equal to E: but AB is supposed to be equal to FE; wherefore, AB is equal to BD; and therefore, (Art. 107.) the angle A is equal

to BDA; again, (Art. 42.) the angles BDC, BDA are equal to two right angles; and the angle E has been shewn to be equal to BDC, and the angle A to BDA: therefore, A and E are together equal to two right angles, which is impossible: for, by the hypothesis, each of them is either greater or less than a right angle. Therefore, AC is equal to EG, and (Art. 86.) the other angles, of ABC, are equal to the remaining angles, of EFG; each to each.

PROP. XX.

(115.) Theorem. If two spherical triangles have two angles of the one equal to two angles of the other, each to each, and the two sides about the third angle of the one, not quadrants, but equal to the two sides about the third angle of the other, each to each, then shall the remaining side be also equal to the remaining side, and the remaining angle to the remaining angle.

In the two spherical triangles ABC*, EFG, let the angles A, C be equal to the angles E, G, each to each; and let the sides AB, BC, which are not quadrants, be equal to EF, FG, each to each: the remaining sides AC, EG are equal, and the remaining angles ABC and EFG are also equal.

For, if AC and EG be unequal, let either of them, as AC, be the greater, and from AC cut off (Art. 92.) CD equal to GE, and join (Art. 66.) B, D.

Then, since BC, CD are equal to FG, GE, each to each, and the angle BCD to the angle FGE, there

See the figure in Art. 114.

fore, (Art. 97.) BD is equal to FE, and the angle BDC to FEG: But, by the supposition, FE is equal to BA, and the angle FEG to BAC: wherefore, BA is equal to BD, and the angle BDC to BAC: also, because BA is equal to BD, the angle BAD, or BAC, is equal (Art. 107.) to BDA; wherefore, BDA is equal to BDC, and (Art. 42.) the angle D, and consequently, also, the angle A, is a right angle: Therefore, (Art. 51.) B is the pole of the circle AC; and (Art. 36.) BA is a quadrant: which is contrary to the hypothesis. Therefore, EG is not unequal to AC; and (Art. 86.) the angles ABC and EFG are equal to one another.

PROP. XXI.

(116.) Theorem. If in two right-angled spherical triangles, having only one angle in each a right angle, the hypotenuses be equal, and if another side, or another angle, not the right angle, in the one, be equal also to another side, or another angle, in the other, the remaining angles and sides shall be equal, in either case, each to each.

Let BAC, FEG be two spherical triangles, having only the angles A and E right angles, and the hypotenuse BC equal to the hypotenuse FG; and, first, let another side, as BA, of BAC, be equal to the side FE, of FEG: then, shall AC be equal to EG, and the other angles, of ABC, to the other angles, of EFG.

For, if AC be unequal to EG, let it be the greater: and make (Art. 92.) AD equal to EG, and join B, D (Art. 66.)