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We show immediately by superposition, and without any preliminary proposition, that two triangles are equal, when a side and the two adjacent angles of the one are equal to a side and the two adjacent angles of the other, each to each. Let us call P the side in question, A and B the two adjacent angles, C the third angle. The angle C then must be entirely determinate, when the angles A and B are known with the side p; for, if several angles C could correspond to the three given things A, B, p, there would be as many different triangles, which would have a side and the two adjacent angles of the one equal to a side and the two adjacent angles of the other, which is impossible; therefore the angle C must be a determinate function of the three quantities A, B, p; which may be expressed


C: (A, B, p).

Let the right angle be equal to unity, then the angles A, B, C, will be numbers comprehended between 0 and 2; and, since

C = (A, B, p),

we say that the line p does not enter into the function p. Indeed we have seen that C must be entirely determined by the data A, B, p, merely, without any other angle or line whatever; but the line p is of a nature heterogeneous to the numbers A, B, C; and if, having any equation whatever among A, B, C, p, we could deduce the value of p, in A, B, C, it would follow that p is equal to a number, which is absurd; therefore p cannot enter into the function, and we have simply C (A, B).....*.

This formula proves already that, if two angles of a triangle are equal to two angles of another triangle, the third must be equal to the third; and, this being supposed, it is easy to arrive at the theorem we have in view.

* It has been objected to this demonstration that if it were applied, word for word, to spherical triangles, it would follow that two known angles would be sufficient to determine the third, which would not be true in this kind of triangles. The answer is, that in spherical triangles there is one element more than in plane triangles, and this element is the radius of the sphere which must not be omitted. Accordingly, let r be the radius; then, instead of having

C='9 (A, B,p), we shall have C = 9 (A, B, p, r), or simply C=9 (A, B,?),

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by the law of homogenous quantities. Now, since the ratio is a number, as well as A, B, C, there is nothing to prevent being found in the fraction 4, and then we can no longer conclude that

C = 9 (A, B).


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Cambridge Math matics.

In the first place let ABC (fig. 274) be a triangle right-angled at Fig. 274. A; from the point A let fall upon the hypothenuse the perpendicu1; lar AD. The angles B and D of the triangle ABD are equal to the angles B and A of the triangle BAC; therefore, according to what has just been demonstrated, the third angle BAD is equal to the third C; for the same reason the angle DAC B; consequently BADDAC, or BAC B+C; but the angle BAC is a right angle; therefore the two acute angles of a right-angled triangle, taken together, are equal to a right angle.

Again, let BAC (fig. 275) be any triangle, and BC a side which is Fig. 275not less than each of the two others; if from the opposite angle A the perpendicular AD be let fall upon BC, this perpendicular will fall within the triangle ABC, and will divide it into two right-angled triangles BAD, DAC. Now in the right-angled triangle BAD the two angles BAD, ABD, are together equal to a right angle; in the right-angled triangle DAC the two angles DAC, ACD, are also equal to a right angle. Consequently the four united, or the three BAC, ABC, ACB, are together equal to two right angles; therefore, in every triangle the sum of the three angles is equal to two right angles.

We see by this that the theorem, considered a priori, does not depend upon a series of propositions, but is deduced immediately from the principle of homogeneity, a principle which exists in every relation among quantities of whatever kind. But we proceed to show that another fundamental theorem of geometry may be deduced from the same source.

The above denominations being preserved, and the side opposite to the angle A being called m, and the side opposite the angle B being called n; the quantity m must be entirely determined by the quantities A, B, p; consequently m is a function of A, B, p, as also SO p'

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=y: (A, B, p). But is a number, as well

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as A and B; therefore the function y must not contain the line


p, and we have simply =y: (A, B), or m = py: (A, B). We


have also in a similar manner n = pw: (A, B).

Let there be another triangle formed with the same angles A, B, C, and having for the opposite sides m', n', p', respectively. Since A and B do not change, we have in this new triangle

m'p' y (A, B),

and n'p'y: (B, A). Therefore m: m' :: n' :: p: p'. Therefore in equiangular triangles the sides opposite to the equal angles are, proportional.

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