PROPOSITION III.

(R) If to the point of intersection A of two great circles, two tangents Ae and Af be drawn; the angle eaf will be the measure of the spherical angle BAC.

For AG is the diameter of the sphere, and ADC and ADB are right-angles, but fAD and CAD are also right-angles, being tangents to the circles ABG and ACG; therefore ƒÃ and ea are parallel to DB and DC, hence the angle eaf is equal to the angle CDB; but CDB is the measure of the spherical angle BAC, therefore eaf is the measure thereof.

In the same manner it may be shewn that the angle ILH is equal to CDB, for it is likewise the inclination of the planes ABG and ACG; hence all spherical angles must be measured on the arc of a great circle 90° distant from the angular point, for CB, and not IH, is the measure of the angle BAC.

PROPOSITION IV.

(S) Any two sides of a spherical triangle are together greater than the third side.

B

C

DEMONSTRATION. Let ABC be the spherical triangle, AB and BC together, are greater than AC.

Since AB, BC, and AC are all arcs of equal circles (L. 130.), if the chords AC, AB, and BC be drawn, they will fall within the sphere, and form a plane triangle ABC, any two sides of which, by plane geometry, are together greater than the third.

Now in equal circles the greater chord cuts off the greater arc, or circumference, therefore any two sides of the spherical triangle ABC, are greater than the third. Q.E.D.

COROLLARY. The shortest distance between any two points on the surface of the sphere, is the arc of the great circle which joins them.

PROPOSITION V.

(T) The three sides of any spherical triangle are together less than the circumference of a circle, or 360°; and any one side is less than a semi-circle, or 180°,

DEMONSTRATION. Let ABC be a spherical triangle, and produce the sides CA and CB to D. Then since all great circles cut each other in two points at a semi-circle's distance (C. 129.),

This is plain from Euclid III. and 2.

A

CAD

CAD and CBD are each of them a semi-circle, therefore CA and CB are each of them less than a semi-circle.

Now AC+CB are greater than AB (S. 132.), and AD+DB are greater than AB.

But AC+CB+AD+DB are equal to the circumference of a circle. Therefore AC+CB+AB are less than the circumference of a circle. Q.E.D.

PROPOSITION VI.

(U) If from the angular points A, B, C, of a spherical triangle ABC as poles, there be described three arcs of circles EF, DE, and DF, forming a new spherical triangle DFE; each side of this new triangle is the supplement of the angle at its pole, viz. the side FE is the supplement of the angle A; DE of the angle B; and DF of the angle c-Likewise each angle of this new triangle is the supplement of that side of the original triangle ABC, to which it is opposite; viz. the angle D is the supplement of BC, the angle F of AC, and the angle E is the supplement of the side AB.

DEMONSTRATION. Since B is the pole of the circle DGHE, every part of this circle is 90° distant from B (H. 130.), therefore the distance between в and E is a quadrant, and BH and BG are quadrants.

Since A is the pole of the circle ELMF, every part of this circle is 90° distant from A (H. 130.), therefore the distance between A and E is a quadrant, and AM and AL are quadrants.

Since A and B have been proved to be equi-distant from E, and AM and BG quadrants, being the side AB produced to M and G; E is the pole of the circle GABM.

In the same manner it may be proved that F is the pole of KALC, and that D is the pole of NBCH.

Therefore EM, EG; FL, FK; DN and DH are quadrants, likewise AM, AL; BH, BG; CN and CK are quadrants; consequently (D. 129.) LM is the measure of the angle A, and NH the measure of the angle D.

Now EM and FL are together equal to a semi-circle, or FL, LE, and LM are together equal to a semi-circle; but FL and LE are equal to FE, therefore FE and LM are together equal to a semicircle; that is they are supplements of each other; but LM is the measure of the angle 4, therefore the angle 4 in

K

F

M

B

H

the

the triangle ABC, and side EF in the triangle DFE, are supplements of each other.

In the same manner it may be proved that DE is the supple. ment of the angle B, and that DF and the angle c are supplements of each other.

It remains now to prove that the angles in the new triangle DFE, and the sides of the original triangle ABC, are supplements of each other.

Since CN and BH are quadrants, they are together equal to a semi-circle, or CN, CH, and BC, are together equal to a semicircle; but NC and CH are equal to NH, therefore NH and BC together make a semicircle; now NH is the measure of the angle D; therefore the angle D in the triangle DFE, is the supplement of the side BC in the triangle ABC.

In the same manner it may be shewn, that the angle F is the supplement of AC, and the angle E of AB.

Q.E.D.

(W) COROLLARY I. The three angles of any spherical triangle ABC are together greater than two right angles, and less than six.

For,

The angle A and the side FE are equal to two right-angles. The angle в and the side DE are equal to two right-angles. The angle c and the side DF are equal to two right-angles. Therefore the three angles A,B,C, together with the three sides FE, DE, and DF, are equal to six right-angles; but the three sides FE, DE, and DF, are together less than four right-angles (T. 132.), therefore the three angles A,B,C, are together greater than two right-angles.

Every spherical angle is less than two right-angles (M. 130.); therefore the sum of any three spherical angles is less than six right-angles.

(X) COROLLARY II. The sides of spherical triangles may be changed into angles, and the contrary.

Thus if three angles of a triangle are given to find the sides; subtract each of the angles from 180°, and the three remainders will be the three sides of a new triangle, with which sides find the angles of the new triangle; then subtract each of these angles from 180°, and the three remainders will be the respective sides of the original triangle, whose angles were given. And the contrary when the sides are given to find the angles.

(Y) COROLLARY III. When the three angles (A,B,C,) of a triangle (ABC) are given to find a side (AC), take the angle (B) opposite to the side required from 180°, and use the remainder and the other two angles (A and c) as sides in a new triangle (DPE): In this new triangle find the angle (P) opposite that side

(DE)

(DE) where the supplement is used. Subtract this angle from 180°, and the remainder will be the side (AC) required.

For if FD and FE be continued till they meet in P; DP and EP being the supplements of FD and FE (T. 132.) are equal to the angles c and A; and the side DE is the supplement of the angle в opposite to AC the side required.

Now find the angle P which is equal to the angle F (Q. 131.) but the supplement of the angle F is equal to the side AC, therefore the supplement of the angle P is equal to the side Ac. If you find the angle PDE it will be equal to the side BC, and PED will be equal to the side AB. For PDE and EDF are supplements of each other, and EDF and BC are supplements of each other, therefore the angle PDE is equal to the side BC.

(Z) COROLLARY IV. A quadrantal triangle may be changed into a right-angled triangle, by calling the supplement of the angle opposite to the quadrantal side, the hypothenuse, and the other angles the legs.

For if AC be a quadrant, the angle at P is a right angle, being the supplement of AC (Y. 134.), and DE is the hypothenuse, being the supplement of the angle B (U. 133.).

PROPOSITION VII.

(A) In any two spherical triangles, if the three sides in the one be equal to the three sides in the other, each to each, the angles which are opposite to the equal sides will be equal.

Let ABC be any triangle on the surface of the sphere. With A as a centre, and the distance AC, describe the small circle CED; with в as a centre, and the distance BC, describe the small circle CFD, crossing CED in the point D. From the points A and B draw the great circles AD and BD, then Ac= AD and BC=BD, by construction, also AB is common to

the two triangles ABC and ADB, therefore the three sides of the one are equal to the three sides of the other, each to each. The angles which are opposite to the equal sides in each triangle are equal. For,

Produce the sides BC and BD to m and to n, so that вm and Bn may be quadrants; through m and n draw the great circle mon. Then since only one * great circle can be drawn through

* For the centre of every great circle is in the centre of the sphere, and three points determine the position of a plane,

two

two given points on the surface of the sphere; if the isosceles triangle oвm, be applied to the isosceles triangle oвn, they will exactly coincide. But no is the measure of the LnBo, and mo =no, is the measure of the Loвm (D. 129.).

Again, produce Ac and AD to the points q and p, so that ap and Aq may be quadrants, and let the great circle prq be drawn; the triangles par and qar will coincide by superposition, therefore the par= Lqar.

In the same manner if DA and DB, and also CA and câ, be extended to quadrants, the angles ADB and CB will coincide, when Ac is placed upon DT, and cs upon DA.

(B) SCHOLIUM. The triangles ACB and ADB, though equal in all their constituent parts, are not superposable, or equal by coincidence *, as Legendre has remarked, Book vii. Prop. 11, of his ÉLÉMENTS de GÉOMÉTRIE, and also in his notes 1st and 12th, where he criticises Dr. Simson. These remarks have led some writers to suppose that demonstrations founded on superposition are nugatory; it is presumed that the above demonstration proves the contrary.

PROPOSITION VIII.

(C) In two triangles, if the three angles in the one be equal to the three angles in the other, each to each, these triangles are equal in all respects.

DEMONSTRATION. The supplements of the three angles in one triangle are equal to the supplements of the three angles in the other and each of these three supplements form a new spherical triangle, whose sides and angles are respectively equal (prop. vi and VII.); and the supplements of the equal angles in the new spherical triangle, are equal to the sides of the original triangle whose angles were given; but the supplements of the angles of the new triangle being equal, the sides of the original triangle are equal; and if the sides be equal, the triangles are equal in all respects. (Prop. VII.) Q. E. D.

PROPOSITION IX.

(D) If there be two sides and the included angle in one spherical triangle, equal to two sides, and the included angle in another, each to each; these two triangles, are equal in all respects.

*This is noticed by Mr. Emerson in the second edition of his Trigonometry, page 146, and likewise by other authors. If the triangles be both situated on the same side of the line AB, they will coincide upon the surface of the sphere. DEMON