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 2 m 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 IKM+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 IM KR), we get IK M not less than R or is not less than the angle KIM. But 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 A B C is not less than 2R, as required. A 29 A 2m 2 A "Cor. Hence, since the sum of the angles of any triangle (A B C), is neither greater nor less than 2 R, it is equal to 2 R, = two right an gles." = 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 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. "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)20, (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=4bc, 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-pn, (4), then a2 = b2 + c2 — 2 n bc, (5); where, since p never passes the limits 0 and 2, a2 (c—b)2 26c - 1, and it is evident by (4) that n cannot pass the limits + 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. 66 = Prop. 2. To find the value of n, that corresponds to the baseangles of any isosceles triangle. "Let ABC be any isosceles triangle; having A B=CB= a, D B a съ E A 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 BCA. Bisect the base of the triangle in E, then draw the right line BE from the vertex B to E, and the triangles A B E, 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 nba, or by reduc. b 2 n a, or n= b=2 na, b AE CE = 2 AB a CB' as required. Also, if we use m instead of n, for the vertical angle (B), we have b2=a2+ a2 — 2 m a a=2 (1 — m) a2, or 1—m= 62 2a2 b=2 n a, we get 2 n2, ... 2 n2 = = 1 — m, or n= ± √13TM, (1), 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 B E 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. "Prop. 3. All right-angled triangles which have one of their acute angles common or equal will have equal values of n corresponding to the common or equal angles. "Let ABE, ABE be two right triangles, right-angled at E and E', and having the common angle A the hypothenuse of the one and leg of the other (which include the common angle A) being in the right lines A G, A F, which include the angle A, that is common to the two triangles; that is, AB and A E' are in A G, A B' and AE in AF. When the angles are equal, but not common, we may imagine A BE to be one of the triangles, and we may suppose the leg of the other triangle that is adjacent to the angle that is equal to A to be applied to A G, so that the angle which equals A shall coincide with A; then will the hypothenuse of the applied triangle lie on AF, and we may conceive that A B' E' represents the applied triangle; so that the case of equal angles is reduced to that of a common angle. Let E CA E from E towards F, and E'CA E' from E' towards G, then draw right lines from C to B, and from C' to B'; and it is evident by Sim., B. I., p. 4, that ABC, AB'C' are isosceles triangles, B E, B'E' being the perpendiculars from their vertical angles to their bases. Join the vertices of the isosceles triangles by the right line B B'=c, and put A B=BC=a, A C=b, b n; also put A B B'C AE CE AB CB = = a', AC = b', = From the triangle A B B' we get, by prop. 1 (by using N, instead of n, to represent the angle A in this triangle, since n represents the angle A in the isosceles triangle A B C), c2 — a2 + a12 — 2 Naa', (1), or if we put N=nx, we get c2: a2+ a2-2n xa a'; and in like manner we get from the triangle B'B C, c2= a2 + d2+2nx'ad, (2); where for we must use + when B' is not between the points A and C, must be used for when B' is between A and C, as is evi = and dent from (1) of prop. 2. Equating the two values of c2, we get, after a slight reduction, a2-d2-2 nxa a 2n x' ad = 0, (3); since 2n a= A C, and a2-d2= (a' + d) (a'd) = (a' + d) X AC (the upper sign being used when B' is not between A and C, and in the contrary case); hence, substituting the values of 2na and a2-d2, by rejecting the common factor A C, (3) is reduced to a'd―xa'x'd=0, or a' (1 — x) + d (1 — x') = 0, or since a'=b+d=A C+CB' (using the upper sign when B' is not between A and C, and when it is between A and C), we get AC(1-x)CB' (1−x + 1-x)=0, (4). Now it is evident that CB' must be arbitrary, and not dependent on AC or 1— X, 1-x'; .. we must have 1-x+1— x'=0, and (4) is reduced to AC (1-x)=0, which, since AC is not = 0, gives 1-x=0, .. 1—x'=0, or x= 1, x' 1; hence (1) and (2) become c2a2 + a2 — 2 n a a', (1′), c2 = a2 + d2 + 2 na d, (2′). In like manner, by regarding the angle A as belonging to the isosceles triangle A B'C', we get from the triangles A B B', B C'B', c2 = a2 + a2 — 2 n'a a', (1"), c2= a22 + d'2 + 2 n'a'd', (2′′); where for we must when B is between A and C, and when B is in AC produced beyond C. By equating the values of c2, as given by (1′) and (1"), we get n' as was to be proved. It is evident that A B E may represent any right triangle having A for one of its acute angles, and its hypothenuse on A G; also ABE may denote any right triangle which has A for one of its acute angles, and its hypothenuse on AF; hence, from what has been shown, n will be the same for all the triangles represented by ABE and ABE'; that is, all right triangles which have a common or equal acute angle will have equal values of n corresponding to the common or equal acute angle. There is one case that apparently forms an exception to what has been shown; and that is when the hypothenuse of a tri use n, or AE AE |