of the sphere, which shall touch the other three." The remaining cases are not more difficult, than those which have been exhibited: and the problem may be solved, generally, exactly in the same manner, as a fourth circle is described, so as to touch three given circles, when all the circles are in the same plane.

PART I.

THE ELEMENTS OF

Spherical Geometry.

SECTION VI.

ON THE COMPARISON OF SPHERICAL SURFACES WITH ONE

ANOTHER.

PROP. I.

(193.) Theorem. Two portions of a sphere's surface, which are contained by equal arches of equal lesser circles, are equal to one another.

For, since the arches, which bound the two figures, are equal to one another, if the one figure be applied to the other, as it evidently may be, so that one of the equal arches which bound it, may coincide with one of the arches bounding the other, the two figures shall wholly coincide: otherwise, three equal circles in a sphere might have a common section; which (Art. 14.) is absurd :

since, therefore, the two figures, when the one is applied to the other, wholly coincide, they are equal to one another.

PROP. II.

(194.) Theorem. Two portions of a sphere's surface, which are contained by the semi-circumferences of great circles, are equal to one another, if the spherical angle, contained by the one pair of semi-circumferences, is equal to the spherical angle contained by the other.

For, if the one figure be applied, as it evidently may be, to the other, so that either of the arches which bound it, may coincide with either of the arches, which bound the other, it is plain that the two figures will wholly coincide, and will, therefore, be equal to one another.

PROP. III.

(195.) Theorem. If the two extremities of any arch of a lesser circle, in a sphere, be joined by the arch of a great circle, and if the two extremities of an equal arch, of an equal lesser circle, be likewise so joined, the two portions of the sphere's surface, thus included, are equal to one another.

For, if the one portion be applied to the other, so that the equal arches of the equal lesser circles shall coincide, the other two boundaries of the figures, namely, the two arches of great circles will also coincide: otherwise, two arches of great circles might pass through two points,

on a sphere's surface, which (Art. 15. and 10.) are not the extremities of a diameter; which (Art. 16.) is absurd. Therefore, the one figure may be placed upon the other, so as wholly to coincide with it: and, consequently, the two figures are equal to one another.

PROP. IV.

1

(196.) Theorem. If two spherical triangles *, having their sides similarly posited, have two sides, of the one, equal to two sides, of the other, each to each, and have, also, the angles contained by those sides equal to one another, they shall be equal.

For it may be shewn, as in E. 4. 1. that the two sides of the one triangle may be made to coincide with the two sides equal to them of the other: therefore (Art. 16.) their bases or third sides, will in that case, coincide; and the two figures, thus wholly coinciding, will be equal to one another.

(197.) COR. Two spherical triangles, having their sides similarly posited †, are also equal, if the three sides

* The spherical figures compared in this and the subsequent propositions, are supposed to be on the same sphere, or on equal spheres, unless the contrary be specified.

Two such triangles may be shewn to be equal, if their sides be not similarly posited; by joining the poles of circles described, about the two triangles, and each of the angular points: from which construction there will result six isosceles spherical triangles, three of which

are

of the one be equal to the three sides of the other, each to each.

For (Art. 86.) the angle contained by any two sides of the one is equal to the angle contained by the two sides equal to them of the other.

PROP. V.

(198.) Theorem. The arches of great circles, which join the extremities of two equal arches of equal and parallel lesser circles, in a sphere, toward the same parts, are equal to one another.

Let AB and CD be two equal arches of the two equal

and parallel circles, AB and CD; and let AC and BD

are equal to the other three, each to each: and the equality of the two given triangles will then plainly appear, by taking equals from equals, or else, by adding equals to equals. But as the same thing follows from a subsequent proposition, the proof, which has been pointed out, is not given at length.