23

A

than And proceeding in the same way from triangle to triangle, until we obtain the mth derived triangle, then the sum of its angles will equal that of the given triangle A B C, and one of its angles will not be greater than 2; where m is of course a positive integer. "Cor. 2. If we obtain the derived triangle whose number is m+1, the sum of two of its angles will equal that angle of the mth triangle which has been shown not to be greater than ; we hence see how from any given triangle to derive another triangle such that the sum of its angles shall equal that of the given triangle, and such that the sum of two of its angles shall not be greater than 2; where A denotes one of the angles of the given triangle.

A

A

"Remark. Cor. 1 is substantially the same as Mr. I. Ivory's process, given at page 189 of the New York edition of J. R. Young's Elements of Geometry.

"Prop. 2. The sum of the angles of any triangle is not greater than two right angles.

A

"Let the triangle A B C, of Prop. 1, represent any triangle, and denote a right angle by R, and if possible let the sum of the angles of the triangle equal 2 A+ V, V being a finite positive angle. Then, using A to represent the angle BAC of the triangle, some positive integer, m, may be found so that the inequality m V A shall exist. From m VA, it follows that V>>, or V is greater than 2m By Cor. 2, Prop. 1, we may derive a triangle from ABC, such that the sum of two of its angles shall not be greater than ; hence the sum of these two angles is less than V; consequently the third angle of the triangle must be greater than 2 R, which is impossible. Hence the sum of the angles of the triangle A B C is not greater than two right angles.

66

A

Prop. 3. The sum of the angles of any triangle is greater than a right angle.

as to make the angles DC F, G FH each equal to the angle BA C, and connect the points D and F, D and G, D and B, by right lines ; also through D draw D K, at right angles to D G. Since A B=DC, AC CF, and the angle BAC equals the angle DCF, the triangles ABC, CDF are identical (Sim., B. I., p. 4), making the sides DF and BC equal to each other, and the angle CDF equal to the angle ABC, and the angle CFD equal to the angle A CB; hence the sum of the angles BAC, BCA equals the sum of the angles BCA, DCF, which equals the sum of the angles CFD, GF H. Since the sum of the three angles at C makes two right angles, and that the sum of the angles at F makes two right angles (Sim., B. I., p. 13), it follows from what has been proved that the angles BC D, DFG are equal, and since B C = DF, CD=FG, the triangles BCD, DFG are identical (Sim., B. I., p. 4), making BD equal to D G, and the angle CBD equal to the angle FDG, and the angle CDB equal to the angle FGD; hence the sum of the angles CB D, CDB equals the sum of the angles CDB, FDG. By Prop. 2, since the sum of the three angles of any triangle is not greater than two right angles, and that the three angles at C make two right angles, it follows that the sum of the angles B and D of the triangle B C D is not greater than the sum of the angles ACB, FCD; hence, and from what has been proved, it follows that the sum of the angles of the triangle ABC is not less than the sum of the angles BDC, CDF, FDG; but it is evident that the sum of these angles exceeds the right angle KDG by a fine angle; hence the sum of the three angles of the triangle ABC exceeds a right angle, as required.

=

"Remark. If A B is extended in the direction A B to E, so that BE BD, and if the points D and E, G and E, are connected by right lines, the point D falls evidently within the pentagonal figure CBEGF; and if R denotes a right angle, the sum of the angles BDC, CDF, FDG, GDE, EDB is equal to 4 R (Sim., B. I., p. 15, Cor. 2). Since the sum of the angles A B D, DBE is equal to 2 R, the sum of the angles at the base of the isosceles triangle B DE is not greater than the angle ABD; consequently the angle BDE is not greater than the angle A BD÷2, which is not greater than half the sum of the angles B D C, CDF, FDG. We now observe that the sum of the angles of the triangle ABC is not less than R+R. For if the sum of the angles of the triangle ABC is not greater than R+R, then by what has been shown the sum of the angles BDC,

CDF, FDG, BDE is not greater than R+R+}+&R= R; consequently the angle D of the triangle EDG is not less than 4 RR = §5 R=2R+, which is impossible; hence the sum of the angles of any triangle, A B C, is greater than R+R, as required.

"Prop. 4. The sum of the angles of any triangle is not less than two right angles.

"If the sum of the angles of any one triangle is not the same as that of any other, then there are evidently some triangles having the sum of their angles less than that of any others; let, therefore, A BC (see fig., Prop. 1) denote a triangle such that the sum of its angles is not greater than that of any other triangle. If R represents a right angle, then, since the sum of the angles of any triangle is greater than R, the sum of the angles of the triangle ABC may be expressed by R+V, V being a positive angle. If we denote the angle BAC of the given triangle by A, then, as in Prop. 2, we may find some positive E integer, m, such that the inequality

A 277%

KH

con

such that the sum of two of its angles, EDF, EFD, shall not be greater than ; .. the sum of these (two) angles is less than V, sequently the third angle E of the triangle is greater than R. Of the two sides DE, EF, let DE be that which is not the less, and through E draw E K, at right angles to DE; then, since the angle DEF is greater than R, the perpendicular will of course meet the side D F at some point, as K, between D and F. Since the sum of the sides DE and E F is greater than D F (Sim., p. 20, B. I.), and that DK is greater than DE (Sim., p. 19, B. I.), we of course have DK greater than KF. Hence extend D F to G, so that G K=DK, and extend EK to H, so that KHE K, and connect G and H

by a right line; also extend E H to I, so that IH= EH, and draw a right line from I to G. Since DK GK, EK HK, and the

=

=

angles DK E, GKH are equal (Sim., Prop. 15, B. I.), the triangles DKE, GKH are identical (Sim., p. 4, B. I.), and DE G H, the angle G H K = the angle DE K= R, and the angle HG K= the angle ED K; hence the triangles EHG, IHG are identical, since they have EH IH, HG common, and that the angles at H are right (Sim., p. 4, B. I.); hence the sides GE, GI are equal, the angle GIH equals the angle G E H, and the angle IG H equals the angle EG H. Since the angle DFE is greater than either of the angles FG E, FEG (Sim., p. 16, B. I.), it follows from what has been shown that the angle E G H is not greater than the sum of the angles E D F, E FD, and of course the sum of the angles E G H, KEF is not greater than V, and since G E F is less than EFD, GEF is not greater than the sum of the angles EG H, HEG is not greater than V+, consequently the sum of the angles of the triangle E GI is not greater than 2V+2. But by hypothesis the sum of the angles of the triangle A B C, which equals the sum of the angles of the triangle D E F, is not greater than the sum of the angles of the triangle EGI; .. 2V+24 is not less than R+V, or V is not less than R- Hence V cannot differ from R by any given angle, as a, so that V=R—a, a being a positive finite angle; for by taking a sufficiently great positive integer for m (which is evidently arbitrary), we shall make less than a, which is absurd ; .. V is not less than R. Hence the sum of the angles of the triangle A B C is not less than 2 R.

2A 2m'

A

A

2m

2 A

2T

2 A

"Cor. Since by Prop. 2 the sum of the angles of any triangle is not greater than 2 R, and from what has been shown in this Prop. it is not less than 2 R, it follows that the sum of the angles of any triangle=2 R= two right angles, as required.

"Appendix to Propositions 3 and 4.

"Lem. No triangle can exist such that the sum of its angles shall be less than any given angle; or such that the sum of its angles shall equal an infinitesimal angle. For, if possible, let A B C be such a tri

angle; then, since the sum of its angles is less than any given angle, each of its angles is of course less than

C

A

B

any given angle. Hence, since the angles C and B are infinitesimal angles, the sides AC and AB must coincide very nearly with the side CB, and .. since AC and AB lie in opposite directions they cannot possibly come near to coincidence with each other; but since the angle A is less than any given angle (or infinitesimal), the sides AC, AB must very nearly coincide with each other and have nearly the same direction, which is absurd. Hence the sum of the angles of a triangle is not less than any given angle (or infinitesimal), but it is a finite quantity, being equal to some finite angle, or the sum of finite angles.

"Remark. By aid of this lemma we are prepared to give a very simple demonstration of Prop. 3.

K M

H

"Prop. 3. The sum of the angles of any triangle is greater than a right angle. Let the triangle A B C, of Prop. 1, denote any triangle, and denote the angle BAC by A, and use V to represent any small finite angle; then we may find some positive integer, m, such that the inequality m>A shall exist, consequently the inequality has place also. Hence, by Cor. 1 and 2 of Prop. 1, we can find a derived triangle, which we shall represent by the triangle HIK, such that the sum of its angles equals that of A B C, and the sum of two of its angles, IHK, IKH, is not greater than is greater than the sum of these two angles. At the point I make the angle HIL, equal to the angle HIK; also make the right line IL equal to the side IK, and draw a right line from H to L; then (Sim., p. 4, B. I.) the triangles HIK, HIL are identical, making the side LH equal to the side HK, the angle IHL equal to the angle IH K, and the angle IL H equal to the angle IKH; hence V is greater than the sum of the four angles IH K, IHL, IKH, ILH. If we connect the points K and L by a right line, it will intersect IH, or IH produced, in some point, M, and the angles at M will be right angles; for the triangles KHM, LHM are identical, since HK=HL, and the side HM common, and that the angle K HM equals the angle LHM, .. the angle KMH equals the angle LMH (Sim., p. 4, B. I.); consequently KM is a perpendicular from the angle K of the triangle HIK to the opposite side HI, or to HI produced (Sim., def. 10, B. I.). We now observe that