the angle A, therefore the measure of the angle a in the triangle ABC, and the measure of the side EE in the triangle DFB, are supplements of each other.

In the same manner it may be proved that DE is the supplement of the measure 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 semi-circle; but NC and CH are equal to NH, therefore NH and BC together make a semi-circle; 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 shown, that the angle F is the supplement of the measure of AC, and the angle E of AB. Q. E. D. (T) 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 B 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 (Z. 138), therefore the three angles A, B, C, are together greater than two right-angles.

Every spherical angle is less than two right-angles (M. 134.); therefore the sum of any three spherical angles is less than six right-angles.

(U) 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.

(W) 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 (▲ and c) as sides in a new triangle (DPE): In this new triangle find the angle (P) opposite that si

(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 are equal to the angles and A; and the side DE is the supplement of the angle B opposite to AC the side required.

Now find the angle P which is equal to the angle F (Q. 135.) but the supplement of the angle r 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.

(X) 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 (W. 137.), and DE is the hypothenuse, being the supplement of the angle B (S. 136.)

PROPOSITION V.

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

DEMONSTRATION. Let ABC be a spherical triangle, any two sides AC and BC, are together greater than AB. Let D be the centre of the sphere, and join DA, Dc, db.

The solid angle at D is contained

by the three plane angles BDC, BDA, CDA, any two of which BDC, CDA, are

D

B

greater than third BDA (Keith's Geom. 1 of X, or EUCLID 20 of XI.)

Now the arc BC is the measure of the angle BDC, the arc ac of CDA, and the arc AB of BDA.

Q. E. D.

Therefore AC and BC together are greater than AB. (Z) COROL. The three sides of a spherical triangle are together less than the circumference of a circle.

For every solid angle is contained by plane angles which are together less than four right angles (Keith's Geom. 2 of X.) and the three sides of the spherical triangle are the measures of a solid angle, and therefore are less than four quadrants, or the circumference of a circle.

PROPOSITION VI. LEMMA.

(A) If two solid angles be composed of three plane angles, such, that the plane angles of the one are equal to the plane

angles of the other, each to each; the planes in which these equal angles are situated, shall have the same inclination to each other.

DEMONSTRATION. Let B and

E be two solid angles, and let the plane angle ABC be equal to the plane angle DEF; ABG to DEI, and GBC to IEF; the angle formed by the planes ABC and ABG is equal to the angle formed by the planes DEF and DEI.

For, suppose the straight line

BG to be elevated above the plane ABC, then from any point G in that line let fall the perpendicular GH upon the plane ABC (Keith's Geom. 11 of IX.) in which plane draw HA and HC perpendiculars to AB and BC, and join ag, gc.

Make EI BG, and from the point I draw IK perpendicular to the plane DEF, in which plane draw KD and KF at right angles to ED and EF, and join ID, IF.

Then because GH is perpendicular to the plane ABC, and that HA is at right angles to AB a straight line in that plane, the angle BAG is a right angle (Keith's Geom. 4 of IX. Corol. 3.): for the same reason EDI is a right angle, therefore the triangle ABG is equal to the triangle DEI.

In the same manner it may be shown that the triangle GBC is equal to the triangle IEF, therefore the quadrilateral figure ABCH, situated in the plane ABC, is equal to the quadrilateral figure DEFK, situated in the plane DEF.

For, if the angle ABC be applied to the angle DEF, ab, BC will coincide with DE and EF; and because all right angles are equal to each other, AH, which is at right angles to AB, will fall upon DK, which is at right angles to ED, and for the same reason CH will fall upon FK, therefore the point н will fall upon the point K.

Now because the triangles AHG and DKI are right-angled at н and x, and the hypothenuse AG is equal to the hypothenuse DI, these triangles are equal (Keith's Geom. 14 of III. note 11.), therefore the angle HAG is equal to the angle KDI.

But the angle HAG is the inclination of the two planes ABC, ABG; and the angle KDI is the inclination of the two planes DEF, DEI; therefore these planes have equal inclinations to each other. Q. E. D.

SCHOLIUM. If the angle HAG be obtuse, viz. if the point н, and the straight line BC fall on contrary sides of AB; the angle KDI will be equally obtuse.

PROPOSITION VII.

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

G

m

C

Let ABC be any triangle on the surface of the sphere.From A as a pole, with the distance AC, describe the small circle cmD; from в as a pole, with the distance BC, describe the small circle CGD, crossing CmD in the point D. From the points A and B draw the great circles AD and BD, then AC AD and BCBD, by construction, also AB is common to the two triangles ABC and ABD, therefore the three sides of the one are equal to the three sides of the other, each to each.

B

The angles which are opposite to the equal sides in each triangle are equal. For,

If o be the centre of the sphere, the solid angle formed at the point o by the three planes AOB, AOD, BOD, will be equal to the solid angle at o formed by the three planes AOB,

AOC, BOC.

Because the sides of the triangles ABC and ABD are equal each to each, by construction; and that these sides are the respective measures of the plane angles which constitute the solid angles at the centre of the sphere (Prop. v. Y. 138.); it follows, that the plane angles which form one of these solid angles, are equal to the plane angles which form the other solid angle, each to each.

But the planes in which such angles are situated have the same inclination to each other (Prop. vi. A. 138.) therefore the angles of the spherical triangle ABC are equal to those of the spherical triangle ABD, viz. DAB= ≤ CAB, DBA≈ ≤ ABC, and ADB ACB.

SCHOLIUM. The equal angles ACB and ADB being situated on contrary sides of the centre of the sphere, the triangle ABC will not coincide with the triangle ADB, unless we conceive the convexity of the one to be applied to the concavity of the other.

If the equal angles ACB and ADB were situated on the same side of the centre of the sphere, the triangle ABC when applied to the triangle ADB would coincide with it.

Triangles which are equilateral, or isosceles, will coincide when applied to each other.

PROPOSITION VIII.

(C) In two triangles, if the three angles of the one be equal to the three angles of 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. IV. 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.) 2. 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.

DEMONSTRATION. Let ABC and EFG be two triangles on the surface of the sphere, and let the side AB EF, AC EG, and the angle BAC FEG. The triangle ABC is equal to the triangle EFG.

First, let the triangles be on the same side of the centre of the sphere. Then if the triangle EFG be applied

B

F

E

to the triangle ABC, EF will coincide with AB and EG with AC (like plane triangles in EUCLID 4 of I.)

The point F coinciding with B, and G with c, the arc FG must coincide with BC, since only one great circle can be drawn through two given points on the surface of the sphere.

Hence, the three sides of the one triangle being equal to the three sides of the other, each to each, the triangles are equal. (Prop. vII.)

Secondly, let the triangles be situated on contrary sides of the centre of the sphere; then a triangle ADB may be constituted contiguous to ABC, and equal to EGF (as in Prop. vII.), and consequently EGF will coincide with ADB. Therefore the side FG BC, the angle EFG ABC, and EGF= 2 ACB. 2. E. D.

For the centre of every great circle is the centre of the sphere, and three given points determine the position of a plane. (Keith's Geom. 2 of IX. Corol.)