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. 2.), let Fig. 23the 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; moreover 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, BDBA + 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, is 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+ IC + IB, that is,

GC+AB <AG+BC.

From one of these take away AB, and from the other its equal
AG, and there remains GC <BC; therefore EF <BC.

Case 11. If the point G (fig. 26) fall upon the side BC, then Fig. 26. it is evident that GC, or its equal EF, is less than BC.

Case III. If the point G (fig. 27) fall within the triangle Fig. 27, BAC, then AG + GC < AB + BC (41), therefore, taking away the equal quantities, AG, AB, we shall have GC < BC, or EF < BC.

THEOREM.

43. Two triangles are equal, when the three sides of the one are equal to the three sides of the other, each to each.

Demonstration. Let the side AB = DE (fig. 23,) AC = DF, Fig. 23′′ BC EF; then the angles will be equal, namely, A = D, BE, and CF.

For, if the angle A were greater than the angle D, as the sides AB, AC, are equal to the sides DE, DF, each to each, the side BC would be greater than EF (42); and if the angle A were less than the angle D, then the side BC would be less than EF; but BC is equal to EF, therefore the angle A can neither be greater nor less than the angle D, that is, it is equal to it. In the same manner it may be proved, that the angle B = E, and that the angle C = F.

44. Scholium. It may be remarked, that equal angles are opposite to equal sides; thus, the equal angles A and D are opposite to the equal sides BC and EF.

THEOREM.

45. In an isosceles triangle the angles opposite to the equal sides are equal.

Demonstration. Let the side AB= AC (fig. 28), then will the Fig. 28. angle C be equal to B.

Draw the straight line AD from the vertex A to the point D the middle of the base BC; the two triangles ABD, ADC, will have the three sides of the one, equal to the three sides of the other, each to each, namely, AD common to both, AB = AC, by hypothesis, and BD = DC, by construction; therefore (43) the angle B is equal to the angle C.

46. Corollary. An equilateral triangle is also equiangular, that is, it has its angles equal.

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 the 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 an gle DCB would be a right angle as well as BCE; and a part would be equal to the whole.

THEOREM.

α

52. If from a point A (fig 31), without a straight line DE, 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.

Fig. 31.