PART FIRST.

OF PLANE FIGURES.

SECTION FIRST.

First Principles, or the Properties of perpendicular, oblique, and parallel Lines.

THEOREM.

27. ALL right angles are equal.

Demonstration. Let the straight line CD be perpendicular to

Fig. 16. AB (fig. 16), and GH to EF, the angles ACD, EGH, will be equal.

Take the four distances CA, CB, GE, GF, equal to each other, the distance AB will be equal to the distance EF, and the line EF may be applied to AB, so that the point E will fall upon A, and the point F upon B. These two lines, thus placed, will coincide with each other throughout; otherwise there would be two straight lines between A and B, which is impossible (25). The point G therefore, the middle of EF, will fall upon the point C, the middle of AB. The side GE being thus applied to CA, the side GH will fall upon CD; for, let us suppose, if it be possible, that it falls upon a line CK, different from CD; since, by hypothesis (10), the angle EGH = HGF,

it follows that

Fig. 17.

But

and

besides, by hypothesis,

hence

and the line GH cannot fall upon a line CK different from CD; consequently it falls upon CD, and the angle EGH upon ACD, and EGH is equal to ACD; therefore all right angles are equal.

THEOREM.

28. A straight line CD (fig. 17), which meets another straight line AB, makes with it two adjacent angles ACD, BCD, which, taken together, are equal to two right angles.

Demonstration. At the point C, let CE be perpendicular to AB. The angle ACD is the sum of the angles ACE, ECD; therefore ACD+ BCD is the sum of the three angles ACE, ECD, BCD. The first of these is a right angle, and the two others are together equal to a right angle; therefore the sum of the two angles ACD, BCD, is equal to two right angles.”

29. Corollary 1. If one of the angles ACD, BCD, is a right angle, the other is also a right angle.

30. Corollary 11. If the line DE (fig. 18) is perpendicular to Fig. 19. AB; reciprocally, AB is also perpendicular to DE.

For, since DE is perpendicular to AB, it follows that the angle ACD is equal to its adjacent angle DCB, and that they are both right angles. But, since the angle ACD is a right angle, it follows that its adjacent angle ACE is also a right angle; therefore the angle ACE = ACD, and AB is perpendicular to DE.

31. Corollary III. All the successive angles, BAC, CAD, DAE, EAF, (fig. 34), formed on the same side of the straight Fig. 34. line BF, are together equal to two right angles; for their sum is equal to that of the two angles BAM, MAF; AM being perpendicular to BF.

THEOREM.

32. Two straight lines, which have two points common, coincide throughout, and form one and the same straight line.

Demonstration. Let the two points, which are common to the two lines, be A and B (fig. 19). In the first place it is evident Fig. 19. that they must coincide entirely between A and B ; otherwise, two straight lines could be drawn from A to B, which is impossible (25). Now let us suppose, if it be possible, that the lines, when produced, separate from each other at a point C, the one becoming CD, and the other CE. At the point C, let CF be drawn, so as to make the angle ACF, a right angle; then, ACD being a straight line, the angle FCD is a right angle (29); and, because ACE is a straight line, the angle FCE is a right angle. But the part FCE cannot be equal to the whole FCD; whence straight lines, which have two points common A and B, cannot separate the one from the other, when produced; therefore they must form one and the same straight line.

Fig. 20.

THEOREM.

33. If two adjacent angles ACD, DCB (fig. 20), are together equal to two right angles, the two exterior sides AC, CB, are in the same straight line.

Demonstration. For if CB is not the line AC produced, let CE be that line produced; then, ACE being a straight line, the angles ACD, DCE, are together equal to two right angles (28); but, by hypothesis, the angles ACD, DCB, are together equal to two right angles, therefore ACD + DCB = ACD + DCE. Take away the common angle ACD, and there will remain the part DCB equal to the whole DCE, which is impossible; therefore CB is the line AC produced.

Fig. 21.

Fig. 22.

THEOREM.

34. Whenever two straight lines AB, DE (fig. 21), cut each other, the angles opposite to each other at the vertex are equal.

Demonstration. Since DE is a straight line, the sum of the angles ACD, ACE, is equal to two right angles; and, since AB is a straight line, the sum of the angles ACE, BCE, is equal to two right angles; therefore ACD+ACE=ACE + BCE; from each of these take away the common angle ACE, and there will remain the angle ACD equal to its opposite angle BCE.

It may be demonstrated, in like manner, that the angle ACE is equal to its opposite angle BCD.

35. Scholium. The four angles, formed about a point by two straight lines which cut each other, are together equal to four right angles; for the angles ACE, BCE, taken together, are equal to two right angles; also the other angles ACD, BCD, are together equal to two right angles.

In general, if any number of straight lines, as CA, CB ( fig. 22), &c., meet in the same point C, the sum of all the successive angles, ACB, BCD, DCE, ECF, FCA, will be equal to four right angles. For, if at the point C, four right angles be formed by two lines perpendicular to each other, they will comprehend the same space as the successive angles, ACB, BCD, &c.

These are often called vertical angles.

THEOREM.

36. Two triangles are equal, when two sides and the included angle of the one are equal to two sides and the included angle of the other, each to each.

Demonstration. In the two triangles ABC, DEF (fig. 23), let Fig. 23. the angle A be equal to the angle D, the side AB equal to the side DE, and the side AC equal to the side DF; the two triangles ABC, DEF, will be equal.

Indeed the triangles may be so placed, the one upon the other, that they shall coincide throughout. If, in the first place, we apply the side DE to its equal AB, the point D will fall upon A, and the point E upon B. But, since the angle D is equal to the angle A, when the side DE is placed apon AB, the side DF will take the direction AC; morcover DF is equal to AC; therefore the point F will fall upon C, and the third side EF will exactly coincide with the third side BC; therefore the triangle DEF is equal to the triangle ABC (26).

37. Corollary. When, in two triangles, these three things are equal, namely, the angle A = D, the side AB = DE, and the side AC= DF, we may thence infer, that the other three are also equal, namely, the angle B = E, the angle C = F, and the side BC= EF.

THEOREM.

38. 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.

Demonstration. Let the side BC (fig. 23) be equal to the side Fig. 23. EF, the angle B equal to the angle E, and the angle C equal to the angle F; the triangle ABC will be equal to the triangle DEF.

For, in order to apply the one to the other, let EF be placed upon its equal BC, the point E will fall upon B and the point F upon C. Then because the angle E is equal to the angle B, the side ED will take the direction BA, and therefore the point D will be somewhere in BA; also because the angle F is equal to C, the side FD will take the direction CA, and therefore the point D will be somewhere in CA; whence the point D, which must be at the same time in the lines BA and CA, can only be at their intersection A; therefore the two triangles ABC,

Fig. 23.

Fig. 24.

Fig. 25.

DEF, coincide, the one with the other, and are equal in all respects.

39. Corollary. When, in two triangles, these three things are equal, namely, BC = EF, B = E, and CF, we may thence infer that the other three are also equal, namely, AB = DE, AC DF, and A = D.

THEOREM.

40. One side of a triangle is less than the sum of the other two. Demonstration. The straight line BC (fig. 23), for example, is the shortest way from B to C (3); BC therefore is less than BA+AC.

THEOREM.

41. If from a point ○ (fig. 24), within a triangle ABC, there be drawn straight lines OB, OC, to the extremities of BC, one of its sides, the sum of these lines will be less than that of AB, AC, the two other sides.

Demonstration. Let BO be produced till it meet the side AC in D; the straight line OC is less than OD + DC; to each of these add BO, and BO + OC < BO + OD + DC; that is

BO + OC <BD + DC.

Again, BD < BA + AD; to each of these add DC, and we shall have BD + DC < BA+AC. But it has just been shown that BO+OC < BD+DC, much more then is

BO+ OC < BA +AC.

THEOREM.

42. If two sides AB, AC (fig. 25), of a triangle ABC, are equal to two sides DE, DF, of another triangle DEF, each to each; if, at the same time, the angle BAC, contained by the former, 19 greater than the angle EDF, contained by the latter; the third side BC of the first triangle, will be greater than the third side EF of the second.

Demonstration. Make the angle CAG D, take AG = DE, and join CG, then the triangle GAC is equal to the triangle EDF (36), and therefore CG EF. Now there may be three cases, according as the point G falls without the triangle ABC, on the side BC, or within the triangle.

Case 1. Because GC GI+IC, and AB < AI + IB, therefore GC+AB < GI + AI + 1C + IB, that is,