In the second figure, CBD and CBA are supplements of each other (M. 134.), and the perpendicular CD falls opposite to both CBD and CAD being of the same species, therefore it cannot fall opposite to CBA at the same time, hence it must necessarily fall without the triangle. Q. E. D.

(X) COROLLARY. In an isosceles triangle, the equal angles are of the same species as their adjacent, or opposite sides.

For then AD DB (K. 143.) and these segments are always acute, since their sum can never be equal to a semi-circle. (C. 133.)

But according as DB and DC, or AD and DC, are of the same, or different species, BC or its equal aC, is acute or obtuse; or since the segments are always acute, BC and AC will be of the same species as DC, but DC is of the same species as A and B, therefore the sides are of the same species as their adjacent or opposite base angles.

PROPOSITION XVI.

(Y) If the LEAST two sides of a spherical triangle be of the same species, the perpendicular drawn from their included angle upon the third side, will fall within the triangle.-Or, if the third side be less than either of the others, they being of the same species, but greater than their supplements, the perpendicular will fall within the triangle, and the less segment of the base and less vertical angle will be adjacent to the greater side, and the greater segment, and greater vertical angle will join the less side, provided the sum of the sides be greater than 180°.

DEMONSTRATION. Let ABC be the triangle, CD a perpendicular upon its base AB; make BN equal to the hypothenuse BC, and AI equal to the hypothenuse ac, then BCN and ACI are isosceles triangles; the angle BNC is of the same species as BC, and the angle aic is of the same species as AC. (X.147.). But AC and BC are of the same species,

INDI

E

by the hypothesis, therefore the angles BNC and AIC are of the same species; consequently CD falls between them (W. 146.) and of course within the triangle ABC.

SECONDLY. In the triangle AVB, where AV and BV are each greater than AB, the least two sides of the supplemental triangle FVE, which by the hypothesis are each less than AB or FE, are of the same species, and consequently va falls between them, therefore VD falls between AV and BV. But Av is greater than BV, and AD is less than DB; consequently AVD is less than BVD. Q. E. D.

(Z) COROLLARY I. The perpendicular let fall on the base of an oblique spherical triangle is either less or greater than each side.

For since CD is perpendicular to AB, it must pass through P the pole of AB (I. 134.); and if c be below the pole P, that is if CD be less than a quadrant, it will be the least of all the lines that can be drawn from the point c to the circumference of the circle ABEF, and CG will be the greatest. (U. 145.)

If c the vertex of the triangle ABC fall above the pole P, suppose in v, then VD will be the perpendicular, and of all the arcs that can be drawn from v upon the arc AB, VD is the greatest (U. 145.), therefore the perpendicular is either less or greater than each side; when it becomes equal to one of the sides, the triangle will be right angled, as BDC.

If the perpendicular fall without the triangle, as in BOC, or BOV, the demonstration will hold equally true.

(A) COROLLARY II. When the perpendicular falls without the triangle, it may be either less or greater than a quadrant; or less or greater than each side.

In the triangle BOC, either CD or CG may be esteemed the perpendicular upon the base OB continued; the former CD is less than a quadrant, and less than either co or CB, and the latter CG is greater.

(B) COROLLARY III. When the perpendicular falls without the triangle, the less perpendicular is next the less side, and the greater perpendicular is next the greater side; and either of them may be considered as the proper perpendicular or the base produced.

In the triangle BOC, CD the less perpendicular is next the less side co, and CG the greater perpendicular is next the greater side BC, and either of them may be used for determining the several parts of the triangle BOC, but the one is sometimes more convenient than the other.

(C) COROLLARY IV. In any triangle ABC or BOC, the perpendicular CD falling within the triangle ABC upon the base AB, or NEAREST perpendicular CD falling without the triangle BOC, is always of the same species as half the sum of the two sides of the triangle; and if half the sum of the two sides of the triangle be acute, this perpendicular falls nearest the less side, if obtuse, it falls nearest the greater side.

In the triangle ABC the perpendicular DC is less than a quadrant, and each side AC and BC is less than a quadrant, there.fore half their sum is less than a quadrant, and the perpendicular CD falls nearest to AC.

In the triangle BOC, CD is less than a quadrant, and each of the sides co and CB is less than a quadrant, therefore half their

sum is less than a quadrant, and the perpendicular falls nearest

to co.

And, in the triangles ABV and BOv, where the perpendicular VD is greater than a quadrant, each of the sides AV, BV, and ov is greater than a quadrant; consequently half their sum must be greater than a quadrant, and the perpendicular vò falls nearest to AV in the former triangle, and to ov in the latter.

(D) COROLLARY V. If an oblique spherical triangle have two acute sides and one obtuse, a perpendicular drawn on the longest side will always fall within the triangle (Y. 147.)-But if it has two obtuse sides and one acute, a perpendicular drawn to fall on the longest side may sometimes fall within, and sometimes without the triangle.

Take any right-angled triangle BDC, and produce its sides to a semi-circle; now because DB and DC are each less than a quadrant, the hypothenuse BC will be less than a quadrant (Q. 145.), consequently its supplement CF will be greater than a quadrant. The angles B and C being acute or each less than a right angle, BC is greater than DB (L. 143.), therefore DF is greater than CF.

Make FA equal to FC, then a great circle as HC may be drawn between A and D; now in the triangle FHC, FH is greater than FC; FC, HC, are of different species, and the perpendicular CD falls without the triangle FHC.

In the triangle Foc, OF is greater than FC; FC and oc are of different species, and the perpendicular CD falls within the triangle Foc.

Therefore there can be no general rule for drawing a perpendicular when the three sides of a triangle are given, especially if two of those sides be obtuse.

(E) COROLLARY VI. If the two obtuse sides of an obliqueangled triangle be EQUAL, a perpendicular drawn from their included angle will fall within the triangle.

For then the triangle is isosceles, and the angles at the base are equal to each other. (F. 142.)

PROPOSITION XVII.

(F) In any spherical triangle, 1. If the three angles be each acute, cach side will be less than a quadrant.

2. If the three angles be each right angles, each side will be a quadrant.

3. If the three angles be each obtuse, each side will be greater than a quadrant.

DEMONSTRATION. First, Let all the

F

A

angles A, B, and c of the triangle ABC be acute, then a perpendicular as CD drawn from any angle will fall within the triangle.

In the right-angled triangle ADC, the angles A and ACD are acute, and therefore (P. and R.145.) AC is less than a quadrant. By the same manner of reasoning from the triangle BDC, BC is less than a quadrant; and because the angles A and abf are acute, AB is less than a quadrant. (P. and R. 145.)

Second. This part is demonstrated (O. 144.)

Third. If the three angles be obtuse, each side of a supplemental triangle will be less than 90° (Prop. vi.); if the three sides are less than 90°, the three angles will be acute, by this proposition, and the supplements of these three angles must necessarily be obtuse; but (U. 137.) they will be the sides of the original triangle, the angles of which were all obtuse. Q. E. D.

(G) COROLLARY I. In any spherical triangle having two obtuse angles and one acute, the sides are of the same species as their opposite angles.

Imagine the triangle AEC to have an acute angle at A, and two obtuse ones at E and c; then the triangle CEG has three acute angles, viz. the angle & is equal to the angle A, (Q. 135.) and the acute angles at E and c are supplements of the obtuse angles in the triangle AFC (M. 184.):

But when the three angles are acute, each side of the triangle CEG is less than a quadrant, therefore AE and AC, opposite to the obtuse angles c and E, are greater than quadrants, and ce opposite to the acute angle a, is less than a quadrant.

(H) COROLLARY II. If a spherical triangle have two sides each less, and one greater than a quadrant; the angles will be of the same species as their opposite sides.

In the triangle AHC, let AH and HC be each of them less than a quadrant, and AC greater; the supplemental triangle to it (U. 137.) will have two obtuse angles and one acute, and consequently the sides thereof, by this proposition, are of the species as their opposite angles : but the supplements of these sides are the angles of the original triangle AHC, therefore it has two acute angles HAC, ACH, and one obtuse, viz. ahc.

same

(I) COROLLARY III. The reverse of this proposition is true in all its parts. Viz. If the three sides be each less than a quadrant, each angle will be acute; if the three sides are each quadrants, the angles will be right-angles; and if the three sides are each greater than quadrants, the three angles will be each obtuse.

These follow from considering the supplemental triangles. (U. 137.)

(K) If two sides of a spherical triangle be of the same species

and the angle included acute, the third side is less than a quadrant; but if the two sides be of different species, and the included angle obtuse, the third side is greater than a quadrant. OR,

When two angles are of the same species, and the included side greater than a quadrant, the third angle is obtuse; but if the two angles be of different species and the included side less than a quadrant, the third angle is acute.

A

JB

D

DEMONSTRATION. Let the triangles BAC and BDC be right-angled at A and D, then will BC be less than a quadrant (Q. 145.); but as the arc ABD approaches nearer to ACD, it will diminish the angles A and D, and cr will consequently be less than CB, which has been proved to be less than a quadrant. In the triangle CID right-angled, suppose at 1, when ID is greater and ic less than a quadrant, the other side, or hypothenuse, CD is greater than a quadrant (Q. 145.); and as the angle DIC increases, viz. when it becomes equal to DIG, the side DC increases till it becomes equal to DG; hence the first part of the proposition is evident. And,

In the second part, the supplemental triangle (U. 137.) will have exactly the same properties as the first part of this proposition. Q. E. D.

PROPOSITION XIX.

E

D

(L) In any right-angled spherical triangle ABC, if the sides be produced, viz. AC to 1, AB to H, and BC to D, so that CI, CD, and AH may be quadrants, or each 90°; and if from the point A, as a pole, the great circle HGFE be described, and from the point c, as a pole, IDE be drawn of such a length that GE may be a quadrant; then shall the triangles CGF and EDF have their respective sides and angles either equal to those of the triangle ABC, or they will be complements of each other; and eight right-angled spherical triangles will be formed, having (every two of them) equal angles at their bases.

H

B

DEMONSTRATION. Since A is the pole of the circle HGFÉ, AG and AH are quadrants and perpendiculars to FGH (H. 133.): and the arcs FGH, FCB, being each of them perpendicular to ABH, are each of them quadrants, and F is the pole of ABH: therefore the triangle CGF is right-angled at G; CG is the com