47. Scholium. From the equality of the triangles ABD, ACD, it follows, that the angle BAD = DAC, and that the angle BDA=ADC; therefore these two last are right angles. Hence a straight line drawn from the vertex of an isosceles triangle, to the middle of the base, is perpendicular to that base, and divides the vertical angle into two equal parts.

In a triangle that is not isosceles, any one of its sides may be taken indifferently for a base; and then its vertex is that of the opposite angle. In an isosceles triangle, the base is that side which is not equal to one of the others.

Fig. 20.

Fig. 30.

THEOREM.

48. Reciprocally, if two angles of a triangle are equal, the opposite sides are equal, and the triangle is isosceles.

Demonstration. Let the angle ABC=ACB (fig. 29), the side AC will be equal to the side AB.

For, if these sides are not equal, let AB be the greater. Take BD AC, and join DC. The angle DBC is, by hypothesis, equal to ACB, and the two sides DB, BC, are equal to the two sides AC, CB, each to each; therefore the triangle DBC is equal to the triangle ACB (36); but a part cannot be equal to the whole; therefore the sides AB, AC, cannot be unequal; that is, they are equal, and the triangle is isosceles.

THEOREM.

49. Of the two sides of a triangle, that is the greater, which is opposite to the greater angle; and conversely, of the two angles of a triangle, that is the greater, which is opposite to the greater side.

Demonstration. 1. Let the angle C > B (fig. 30), then will the side AB, opposite to the angle C, be greater than the side AC, opposite to the angle B.

Draw CD making the angle BCD=B. In the triangle BDC, BD is equal to DC (48); but AD + DC > AC, and

AD+DC=AD+ DB = AB, therefore AB>AC. 2. Let the side AB> AC, then will the angle C, opposite to the side AB, be greater than the angle B, opposite to the side AC. For, if C were less than B, then according to what has just been demonstrated we should have ABAC, which is contrary to the hypothesis; and if C were equal to B, then it would

follow, that AC = AB (48), which is also contrary to the hypothesis; whence the angle C can be neither less than B, nor equal to it; it is therefore greater.

THEOREM.

50. From a given point A (fig. 31), without a straight line DE, Fig 31. only one perpendicular can be drawn to that line.

Demonstration. If it be possible, let there be two AB and AC; produce one of them AB, so that BF = AB. and join CF.

=

The triangle CBF is equal to the triangle ABC. For the angle CBF is a right angle (29), as well as CBA, and the side. BF = BA; therefore the triangles are equal (36), and the angle BCF BCA. But BCA is, by hypothesis, a right angle; therefore BCF is also a right angle. But, if the adjacent angles BCA, BCF, are together equal to two right angles, ACF must be a straight line (33); and hence it would follow that two straight lines ACF, ABF, might be drawn between the same two points A and F, which is impossible (25); it is then equally impossible to draw two perpendiculars from the same point to the same straight line.

51. Scholium. Through the same point C (fig. 17), in the Fig. 17. line AB, it is also impossible to draw two perpendiculars to that line; for, if CD and CE were these two perpendiculars, the angle DCB would be a right angle as well as BCE; and a part would be equal to the whole.

THEOREM.

52. If from a point A (fig. 31), without a straight line DE, a Fig. 31 perpendicular AB be drawn to that line, and also different oblique lines AE, AC, AD, &c., to different points of the same line;

1. The perpendicular AB is less than any one of the oblique lines;

2. The two oblique lines AC, AE, which meet the line DE on opposite sides of the perpendicular, and at equal distances BC, BE, from it, are equal to one another;

3. Of any two oblique lines AC, AD, or AE, AD, that which is more remote from the perpendicular is the greater.

Demonstration. Produce the perpendicular AB, so that BF = BA, and join FC, FD.

1. The triangle BCF is equal to the triangle BCA; for the right angle CBF = CBA, the side CB is common, and the side BF BA; therefore the third side CF is equal to the third side AC (36). But AF < AC+ CF (40), and AB half of AF is less than AC half of AC + CF, that is, the perpendicular is less than any one of the oblique lines.

2. If BE BC, then, as AB is common to the two triangles ABE, ABC, and the right angle ABE = ABC, the triangle ABE is equal to the triangle ABC, and AE = AC.

3. In the triangle DFA, the sum of the sides AD, DF, is greater than the sum of the sides AC, CF (41); therefore AD half of AD + DF is greater than AC half of AC + CF, and the oblique line, which is more remote from the perpendicular, is greater than that which is nearer.

53. Corollary 1. The perpendicular measures the distance of any point from a straight line.

54. Corollary 1. From the same point, there cannot be drawn three equal straight lines terminating in a given straight line; for, if this could be done, there would be on the same side of the perpendicular two equal oblique lines, which is impossible.

Fig. 32.

THEOREM.

55. If from the point C (fig. 32), the middle of the straight line AB, a perpendicular EF be drawn; 1. each point in the perpendicular EF is equally distant from the two extremities of the line AB; 2. any point without the perpendicular is at unequal distances from the same extremities A and B.

Demonstration. 1. Since AC CB, the two oblique lines AD, DB, are drawn to points which are at the same distance from the perpendicular. They are therefore equal (52). The same reasoning will apply to the two oblique lines AE, EB, also to AF, FB, &c. Whence each point in the perpendicular EF is equally distant from the extremities of the line AB.

2. Let I be a point out of the perpendicular; join IA IB, one of these lines must cut the perpendicular in D; join DB, then DB DA. But the line IB < ID + DB and

ID+DB = ID +DA = IA ;

therefore IB <IA; that is, any point without the perpendicular is at unequal distances from the extremities of AB.

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 AC DF (fig. 33), and Fig. 38. 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, DF = AC; 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.

THEOREM.

57. If two straight lines, AC, BD, (fig. 35), are perpendicular to Fig. 35. a third AB, these two lines are parallel, that is, they will not meet, however far they are produced (12).

Demonstration. If they could meet in a point O on one side or the other of the line AB, there would be two perpendiculars drawn from the same point O to the same straight line AB, which is impossible (50).

LEMMA.

58. The straight line BD (fig. 35), being perpendicular to AB, Fig. 35: if another straight line AE make with AB an acute angle BAE, the straight lines BD, AE, being produced sufficiently far, will meet.

Demonstration. From any point F, taken in the direction AE, let fall upon AB the perpendicular FG; the point G will not fall upon A, since the angle FAB is less than a right angle; still less can it fall upon H in BA produced, for then there would be two perpendiculars KA, KH, let fall from the same point K upon the same straight line AH. The point G then must fall, as the figure represents it, in the direction AB.

Let there be taken in the line AE another point L at a distance AL greater than AF, and let there be drawn to AB the perpendicular LM; it may be shown, as in the preceding case, that the point M can neither fall upon G nor upon any point in the direction GA, so that the distance AM will be necessarily greater than AG.

It may be observed, moreover, that if the figure is constructed with care, and AL be taken double of AF, we shall find that AM is exactly double of AG; also, if AL be taken triple of AF, we shall find that A is triple of AG, and in general there will always be the same ratio between AM and AG, that there is between AL and AF. From this proportion it follows, not only that the right line AE, must meet BD, if the two lines are produced sufliciently far, but also that we may even assign upon AE the distance of the point of meeting of the two lines. This distance will be the fourth term of the proportion,

AG: AB:: AF: x.

59. Scholium. The foregoing explanation, founded upon a relation which is not deduced from reasoning merely, and for which recourse is had to measures taken upon a figure accurately constructed, has not the same degree of strictness, as the other demonstrations of elementary geometry. It is given here only as a simple method, by which one may satisfy himself of the truth of the proposition. We shall resume the subject with a view to a rigorous demonstration in the third of the notes subjoined to these elements.

Fig. 36.

THEOREM.

60. If two straight lines AC, BD, (fig. 36), make with a third AB two interior angles CAB, ABD, the sum of which is equal to two right angles, the two lines AC, BD, are parallel.

Demonstration. From the point G, the middle of AB, draw the straight line EGF perpendicular to AC; this line will be perpendicular to BD. Indeed the sum GAE + CBD is, by hypothesis, equal to two right angles, and the sum GB+GBD is also equal to two right angles (28); taking therefore from each GBD we shall have the angle GAE = GBF. Besides, the angles AGE, BGF, are equal, being vertical angles; therefore the triangles AGE, BGF, have a side and the two adjacent angles of the one res