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THEOREM.

56. Two right-angled triangles are equal, when the hypothenuse and a side of the one are equal to the hypothenuse and a side of the other, each to each.

Demonstration. Let the hypothenuse ACDF (fig. 33), and Fig. 33. the side AB = DE; the right-angled triangle ABC will be equal to the right-angled triangle DEF.

The proposition will evidently be true, if the third side BC be equal to the third side EF. If it be possible, let these sides be unequal, and let BC be the greater. Take BG = EF, and join AG; then the triangle ABG is equal to the triangle DEF, for the right angle B is equal to the right angle E, the side AB = DE, and the side BG = EF; therefore, these two triangles being equal (36), AG = DF; and, by hypothesis, DFAC; whence AG AC. But AG cannot be equal to AC (52); therefore it is impossible that BC should be unequal to EF, that is, it is equal to it, and the triangle ABC is equal to the triangle DEF.

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THEOREM.

57. In any triangle, the sum of the three angles is equal to two right angles.

Demonstration. Let ABC (fig. 35) be the proposed triangle, Fig. 35 in which we suppose* that AB is the greatest side, and BC the least, and that, consequently, ACB is the greatest angle, and BAC the least (49).

Through the point A, and the middle point I of the opposite side BC, draw the straight line AI, and produce it to C', making ACAB; produce also AB to B', making AB' double of AI.

If we designate by A, B, C, the three angles of the triangle ABC, and by A', B', C', the three angles of the triangle A B'C', we say that CB+ C, and A= A + B'; from which we deduce A+B+C=A'+B+C'; that is, the sum of the three angles is the same in the two triangles.

To prove this, make AK = AI, and join C' K; we shall have

* This supposition does not exclude the case in which the mean side AC is equal to one of the extremes AB or BC.

the triangle C'AK BAI. For, in these two triangles, the angle A is common, and the side ACAB, and AKAI. Therefore the third side C'K is equal to the third BI, and consequently the angle AC'K= ABC, and the angle AKC' = AIB.

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We say now that the triangle B'C'K is equal to the triangle ACI, since the sum of the two adjacent angles AKC + C'KB' is equal to two right angles (28), as well as the sum of the two angles AIC+AIB; subtracting from these, respectively, the equal angles AKC', AIB, and there will remain the angle C'KB'AIC. These equal angles in the two triangles are comprehended between sides that are equal, each to each, namely, C'K=IB= CI, and KB' — AKAI, since we have supposed AB' = 2 AI = 2AK. Therefore the two triangles B'C'K, ACI, are equal (36), and, consequently, the side C'B' = AC, and the angle B'C'K=ACB, and the angle KB'C' = CAI.

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It hence follows, 1. that the angle AC'B', designated by C', is composed of two angles, equal, respectively, to the two angles B and C, of the triangle ABC, and that, accordingly, we have C'=B + C ; 2. that the angle A of the triangle ABC is composed of the angle A', or CA'B', which belongs to the triangle AB'C', and the angle CAI, equal to B', of the same triangle, which gives A=A' + B'; therefore A+B+C A+B'+C'. Moreover, since, by hypothesis, we have AC< AB, and, consequently, C'B' <AC', it will be seen, that, in the triangle AC'B', the angle at A, designated by A', is less than B'; and, as the sum of the two is equal to the angle A of the proposed triangle, it follows that the angle A'< į A.

If we apply the same construction to the triangle AB'C', in order to form a third triangle AC"B", designating the angles by A", B", C", respectively, we shall have, in like manner, the two equations C"C'B', and A'=A′′ +B", which gives A' + B'+C'=A′′ +B′′ +C". Thus the sum of the three angles is the same in these three triangles. We have, at the same time, the angle A"<A', and, consequently, A" < ¦ A.

Continuing indefinitely the series of triangles AC'B', AC"B", &c., we shall arrive at a triangle a b c, in which the sum of the three angles will always be the same as in the proposed triangle ABC, and which will have the angle a less than any given term of the decreasing progression A, A, † A, &c.

We may therefore suppose this series of triangles continued until the angle a is less than any given angle.

Accordingly, if, by means of the triangle abc, we construct the following triangle a'b'c', the sum of the angles a+b' of this triangle will be equal to the angle a, and will, consequently, be less than any given angle; it will hence be seen that the sum of the three angles of the triangle a'b'c' reduces itself to the single angle c'.

In order to obtain the exact measure of this sum, let us produce the side a'c' toward d', and designate the exterior angle b'c'd' by x'; this angle x', added to the angle c' of the triangle a'b'c', will make a sum equal to two right angles (28); thus, denoting the right angle by D, we shall have c'2 D-x'; therefore the sum of the angles of the triangle a'c'b' will be 2D+a+b'x'.

But we may imagine the triangle a'c'b' to vary in its angles and sides, so as to represent the successive triangles which are derived ultimately from the same construction, and which approach more and more the limit at which the angles a' and b' are nothing. At this limit, the straight line a'c'd' is confounded with a'b', and the three points a', c', b', are in the same straight line; then the angles b' and x' become nothing at the same time with a', and the quantity 2 D+ a + b'x', which is the measure of the three angles of the triangle a'c'b', reduces itself to 2 D; therefore, in any triangle, the sum of the three angles is equal to two right angles.

58. Corollary 1. Two angles of a triangle being given, or only their sum, the third will be known by subtracting the sum of these angles from two right angles.

59. Corollary II. If two angles of one triangle are equal to two angles of another triangle, each to each, the third of the one will be equal to the third of the other, and the two triangles will be equiangular.

60. Corollary III. In a triangle, there can be only one right angle; for if there were two, the third angle must be nothing. Still less, then, can a triangle have more than one obtuse angle.

61. Corollary IV. In a right-angled triangle, the sum of the acute angles is equal to a right angle.

62 Corollary v An equilateral triangle, as it must be al

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so equiangular (45), has each of its angles equal to a third of two right angles; so that, if a right angle be expressed by unity, the angle of an equilateral triangle will be expressed by 2.

63. Corollary vi. In any triangle ABC, if we produce the side AB toward D, the exterior angle CBD will be equal to the sum of the two opposite interior angles A and C; for, by adding to each the angle ABC, the sums are each equal to two right angles.

THEOREM.

64. The sum of all the interior angles of a polygon is equal to as many times two right angles as there are units in the number of sides minus two.

Demonstration. Let ABCDE, &c. (fig. 42), be the proposed polygon; if, from the vertex of the angle A, we draw to the vertices of the opposite angles the diagonals AC, AD, AE, &c., it is evident, that the polygon will be divided into five triangles, if it have seven sides, and into six, if it have eight, and, in general, into as many triangles, wanting two, as the polygon has sides; for these triangles may be considered as having for their common vertex the point A, and for their bases the different sides of the polygon, except the two which form the angle BAG. We see, at the same time, that the sum of the angles of all these triangles does not differ from the sum of the angles of the polygon; therefore this last sum is equal to as many times two right angles, as there are triangles, that is, as there are units in the number of sides of the polygon minus two.

65. Corollary 1. The sum of the angles of a quadrilateral is equal to two right angles multiplied by 4-2, which makes four right angles; therefore, if all the angles of a quadrilateral are equal, each of them will be a right angle, which justifies the definition of a square and rectangle (17).

66. Corollary II. The sum of the angles of a pentagon is equal to two right angles multiplied by 5-2, which makes 6 right angles; therefore, when a pentagon is equiangular, each angle is equal to a fifth of six right angles, or g of one right angle.

67. Corollary III. The sum of the angles of a hexagon is equal to 2 × (6—2), or 8, right angles; therefore, in an equi

angular hexagon, each angle is the sixth of eight right angles, or of one right angle. The process may be easily extended to other polygons.

68. Scholium. If we would apply this proposition to polygons, which have any re-entering* angles, each of these angles is to be considered as greater than two right angles. But, in order to avoid confusion, we shall confine ourselves in future to those polygons, which have only saliant angles, and which may be called convex polygons. Every convex polygon is such, that a straight line, however drawn, cannot meet the perimeter in more than two points.

THEOREM.

69. If two straight lines AB, CD (fig. 36), are perpendicular Fig 36. to a third FG, these two lines will be parallel, that is, they will not meet if produced ever so far.

Demonstration. For, if they should meet in any point 0, there would be two perpendiculars OF, OG, let fall from the same point O upon the same straight line FG, which is impossible (50).

THEOREM.

70. If two straight lines AB, CD (fig. 36), make, with a third Fig. 36. EF, two interior angles BEF, DFE, the sum of which is equal to two right angles, the lines AB, CD, will be parallel.

Demonstration. If the angles BEF, DFE, are equal, they will each be right angles, and we fall upon the case of the preceding proposition. Let us suppose, then, that they are unequal; and that, through the point F, the vertex of the greater, we let fall upon AB a perpendicular FG.

In the triangle EFG, the sum of the two acute angles FEG + EFG is equal to a right angle (61); this sum being subtracted from the sum BEF + DFE, equal by hypothesis to two right angles, there will remain the angle DFG equal to a right angle. Therefore the two lines AB, CD, are perpendicular to the third line FG, and are consequently parallel (69).

* A re-entering angle is one whose vertex is directed inward, as Fig. 43. CDE (fig. 43), while a saliant angle has its vertex directed outward, as ABC.

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