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by a right line; also extend EH to I, so that IH= E H, and draw a right line from I to G. Since DK GK, EK HK, and the angles D KE, GKH are equal (Sim., Prop. 15, B. I.), the triangles DKE, G KH are identical (Sim., p. 4, B. I.), and DE = GH, the angle G H K the angle DEK= R, and the angle H G K= the angle ED K; hence the triangles E HG, IHG are identical, since they have EH IH, HG common, and that the angles at. Hare right (Sim., p. 4, B. I.); hence the sides G E, 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 EGH, KEF is not greater than V, and since G E F is less than EF D, GEF is not greater than 2. the sum of the angles EGH, HEG is not greater than V+, consequently the sum of the angles of the triangle E GI is not greater than 2V+ 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

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2 A 2m'

A

A

24

2m

2 A
2 m.

angle, as a, so that VR-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 ABC is not less than 2 R.

2

"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.

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"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

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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

"Prop. 3. The sum of the angles of any triangle is greater than a right angle. Let the triangle AB 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. >> 2m 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

A

than the sum of these two angles.

..

At the point I

H

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 KHM equals the angle L HM, .. 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

KM must fall without the triangle HIK (or that it will meet HI, produced in the direction HI), for if KM does not fall without the triangle KIH, but coincides with K I, or falls at some point between H and I, then we shall have the triangle H K L such that the sum of its angles is less than V, and as V is any finite angle taken as small as we please,.. the sum of the angles of the triangle HKL is less than any finite angle, or it is infinitesimal, which is impossible. Hence the perpendicular K M falls on HI produced in the direction HI, so as to make the angle HKM equal to some finite angle; and it is evident that the perpendicular cannot intersect HI produced in the direction IH, for if it could, a triangle would be formed having the sum of two of its angles greater than two right angles, which is impossible. Hence, since the angle HIK is the exterior angle of the triangle KIM, it is greater than the right angle KMI (Sim., p. 16, B. I.); hence the sum of the angles of any triangle is greater than a right angle, as required.

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

A 29

A

"We shall use the figure to Prop. 3'. It is evident that we may suppose the sum of the angles of the triangle KHL not less than that of the triangle KIH, or, since the sum of the angles IHK, IKH is not greater than we shall have 2 IK M+2, not less than the angle KIH. Hence, if we denote a right angle by R, since the sum of the angles HIK, KIM is equal to 2R (Sim., p. 13, B. I.), and that the sum of the angles KIM, IKM is not greater than R (see our Prop. 2, and observe that the angle IMKR), we get IK M not less than R- or is not less than the angle KIM. But 2m is less than any given angle, .. the angle KIM is infinitesimal, consequently the angle KIH differs from 2R by an infinitesimal angle, and of course the sum of the angles of the triangle KIH or ABC is not less than 2R, as required.

A

A

A

2m

"Cor. Hence, since the sum of the angles of any triangle (ABC), is neither greater nor less than 2 R, it is equal to 2 R, two right angles."

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II. "An attempt to show (analytically) that the sum of the angles of any rectilineal triangle is equal to two right angles.

"Ax. The angle formed by two (right) lines is independent of the lengths of the lines.

"Prop. 1. To express any side of a triangle in terms of the other sides and their included angle.

"Let ABC be any triangle; and suppose its sides B C, A C, A B severally contain some assumed length (considered as the unit of

B

A

B

length) a, b, c times, then will the sides be expressed by a, b, c ; where it may be observed that a, b, c are positive, and that they may be integral or fractional, rational or surd, according to the nature of the case; we shall denote the angle BAC by A, and shall suppose CA to be produced (in the direction CA) to B', so that AB' = BA, then (Simson's Euclid, Book I., prop. 13) the angle BA B' is the supplement of A, or the sum of A and BA B' is equal to two right angles.

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"By Sim., B. I., p. 20, we have the inequalities a+b>c, a+c>b, or (which is equivalent to them), we have a>±(c—b), (1); in which we must use the upper sign when c is greater than b, and the lower sign must be taken when c is less than b; and it is manifest that (1) exists even when c = b. In order to remove the ambiguous sign, we may (by taking the second power of a, and ±(c—b) put (1) under the form a> (c-b)2, or a2 (c—b)2>0, (2). If the angle A0, AB falls on AC, and (2) evidently becomes a2 - (c—b)2=0, which is its least value; and, Sim., B. I., p. 24, if we suppose b and c each invariable, and the angle A to be increased, then A will be increased, and the greatest value that a can have will be when the angle A equals two right angles, or when AB coincides with AB', and a = bc, so that (2) becomes (b+c)2 - (b−c)2=4b c, which is its greatest value; hence and by (2) if we put p, (3), p cannot be less than 0 (or cannot be negative), nor greater than 2, or p has 0 for its lesser, and 2 for its greater limit. From (3) we get a2=(c-b)2 + 2 p b c — b2 + c2 — 2 (1 − p) b c, or if we put 1-p=n, (4), then a2 = b2 + c2 — 2n bc, (5); where, since p never passes the limits 0 and 2,

a2 - (c—b)2
26c

=

it is evident by (4) that n cannot pass the limits + 1 and 1, and that n depends on the angle A; also that n=1 corresponds to A=0, and n=- - 1 to A= two right angles; so that a is expressed in (5) as required.

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Prop. 2. To find the value of n, that corresponds to the baseangles of any isosceles triangle.

"Let ABC be any isosceles triangle; having AB=

for its sides, and A C=b for its base, and let the base be produced in the direction AC to any point, D, then, Sim., B. I., p. 13, the angle BCD is the

supplement of the angle D

ВСА. Bisect the base

B

CB=a,

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of the triangle in E, then draw the right line BE from the vertex B to E, and the triangles A BE, CBE are identical (Sim., B. I., pp. 8 and 4); so that the angle A E B equals the angle CE B, and these angles are right, Sim., B. I., def. 10, and BE is perpendicular to the base of the triangle. By (5) of prop. 1, we get a2: a2 + b2 - 2 n ba, or by reduc. b=2 n a, or n=

b AE

AB

a

=

CE

CB'

as required. Also, if we use m instead of n, for the vertical angle (B), we

have b2=a2+ a2· 2 maa=

62 2a2

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= 2 (1 — m) a2, or 1—m=

12

2a29

or since

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b=2 na, we get 2 n”, . 2 n which is another form of n; and it is manifest that if we take the upper sign before the radical for the value of n that corresponds to the acute angle B CA, we must take the lower sign before the radical in order to get the value of n that corresponds to the obtuse angle BCD, which (as before noticed) is the supplement of BCA (BAC).

"Cor. 1. By what has been done it is evident, that, if we divide one of the legs of a right-angled triangle by the hypothenuse, we get the value of n that corresponds to the included angle; for evidently the same value of n corresponds to the isosceles triangle A B C, and to the identical right triangles into which it is divided by the perpendicular BE from its vertical angle.

"Cor. 2. It is manifest from (1), that all those isosceles triangles which have equal values of n for their base angles also have equal values of m for their vertical angles.

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