Fig. 37. 64. If two parallel straight lines AB, CD (fig. 37), meet a third line EF, the sum of the interior angles upon the same side AGH, GHC, will be equal to two right angles. Demonstration. If this sum were greater or less than two right angles, the two straight lines AB, CD, would meet on one side or the other of EF, and would not be parallel (61). 65. Corollary i. If GHC be a right angle, AGH will also be a right angle; therefore every line, which is perpendicular to one of the parallels, is also perpendicular to the other. 66. Corollary 11. Since the sum AGH+GHC is equal to two right angles, and the sum GHD + GHC is also equal to tworight angles, if we take away the common part GHC, we shall have the angle AGH= GHD. Besides, AGH=BGE, and GHD= CHF (34); therefore the four acute angles AGH, BGE, GHD, CHF, are equal to each other; the same may be proved with respect to the four obtuse angles AGE, BGH, GHC, DHF. It may be observed, moreover, that by adding one of the four acute angles to one of the four obtuse angles, the sum will always be equal to two right angles. 67. Scholium. The angles of which we have been speaking, compared, two and two, take different names. We have already called the angles AH, GHC, interior upon the same side; the angles BGH, GHD, have the same name; the angles AGH, GHD, are called alternate-internal, or simply alternate; the same may be said of the angles BGH, GHC. Lastly, we denominate internal-external the angles EGB, GHD, and EGA, GHC, and alternate-external EGB, CHF, and AGE, DHF. This being premised, we may regard the following propositions as already demonstrated. 1. The two interior angles upon the same side, taken together, are equal to two right angles. 2. The alternate-internal angles are equal, as also the internalexternal, and the alternate-external. Reciprocally, if in this second case, two angles of the same name are equal, we may infer that the lines to which they are referred are parallel. Let there be, for example, the angle AGH= GHD; since GHC + GHD is equal to two right angles, THEOREM. we have also AGH + GHC equal to two right angles, therefore the lines AG, CH, are parallel (60). 24 68. Two lines AB, CD (fig. 38), which are parallel to a third Fig. 36. EF, are parallel to one another. Demonstration. Draw PQR perpendicular to EF. Then, since AB is parallel to EF, the line PR will be perpendicular to AB (65); also, since CD is parallel to EF, the line PR will be perpendicular to CD. Consequently AB and CD are perpendicular to the same straight line PQ, therefore they are par. allel (57). THEOREM. 69. Two parallel lines are throughout at the same distance from each other. Demonstration. The two parallels AB, CD (fig. 39), being Fig. 39. given, if through two points taken at pleasure we erect upon AB the two perpendiculars EG, FH, the straight lines EG, FH, will be at the same time perpendicular to CD (65); moreover these straight lines will be equal to each other. For, hy drawing GF, the angles GFE, FGH, considered with reference to the parallels AB, CD, being alternate-internal angles (67), are equal; also since the straight lines EG, FH, are perpendicular to the same straight line AB and consequently parallel to each other, the angles EGF, GFH, considered with reference to the parallels EG, FH, being alternate-internal angles, are equal. The two triangles then EFG, FGH, have a side and the two adjacent angles of the one equal to a side and the two adjacent angles of the other, each to each; these two triangles are therefore equal (38); and the side EG, which measures the distance of the parallels AB, CD, at the point E, is equal to the side FH, which measures the distauce of the same parallels at the point F. 2 THEOREM. 70. If two angles, BAC, DEF (fig. 40), have their sides par- Fig. 40. allel, each to each, and directed the same way, these two angles will Demonstration. Produce DE, if it be necessary, till it meet AC in G; the angle DEF is equal to DGC, because EF is parallel to GC (67); the angle DGC is equal to BAC, because DG is parallel to AB; therefore the angle DEF is equal to BAC. 71. Scholium. There is a restriction in this proposition, namely, that the side EF be directed the same way as AC, and ED the same way as AB; the reason is this; if we produce FE toward H, the angle DEH would have its sides parallel to those of the angle BAC, but the two angles would not be equal. In this case the angle DEH and the angle BAC would together make two right angles. 27 THEOREM. Fig. 41. 72. In every triangle the sum of the three angles is equal to two right angles. Demonstration. Let ABC (fig. 41) be any triangle; produce the side CA toward D, and draw to the point A the straight line AE parallel to BC. Because AE, CB, are parallel, the angles ACB, DAE, considered with reference to the line CAD, are equal, being internalexternal angles (67); in like manner ABC, BAE, considered with reference to the line AB, are equal, being alternate-internal angles; consequently the three angles of the triangle ABC make the same sum as the three angles CAB, BAE, EAD; therefore this sum is equal to two right augles (31). 73. 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. 74. Corollary 11. 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. 75. Corollary 111. 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. 76. Corollary iv. In every right-angled triangle the sum of the acate angles is equal to a right angle. 77. Corollary v. Every equilateral triangle, as it must be also 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 š. 78. Corollary vi. In every triangle ABC (fig. 41) the exterior Fig.41, angle BAD is equal to the two opposite interior angles B and C'; for, AE being parallel to BC, the part BAE is equal to the angle B, and the other part DAE is equal to the angle C (67). THEOREM. 79. 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 Fig. 42. 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, 80. Corollary i. 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). 81. 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 of one right angle. 82. Corollary in. The sum of the angles of a hexagon is equal to 2 x (6 — 2), or 8, right angles; therefore, in an equiangular hexagon, each angle is the sixth of eight right angles, or of one right angle. The process may be easily extended to other polygons. 83. Scholium. If we would apply this proposition to polygons, which have any re-enteringt 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. 2 THEOREM. Fig. 44. 84. The opposite sides of a parallelogram are equal, and the opposite angles also are equal. Demonstration. Draw the diagonal BD (fig. 44); the two triangles ADB, DBC, have the side BD common; moreover, on account of the parallels AD, BC, the angle ADB = DBC (67), and on account of the parallels AB, CD, the angle ABD=BDC; therefore the two triangles ADB, DBC, are equal (38); consequently the side AB opposite to ADB is equal to the side DC opposite to the equal angle DBC, and likewise the third side AD is equal to the third side BC; therefore the opposite sides of a parallelogram are equal. Again, from the equality of the same triangles it follows, that the angle A= C, and also that the angle ADC, composed of the two angles ADB, BDC, is equal to the angle ABC, composed of the two angles DBC, ABD; therefore the opposite angles of a parallelogram are equal. 85. Corollary. Hence two parallels AB, CD, comprehended between two other parallels AD, BC, are equal. 30 86. If, in a quadrilateral ABCD (fig. 44), the opposite sides are equal, namely, AB = CD, and AD = CB, the equal sides will be parallel, and the figure will be a parallelogram. Demonstration. Draw the diagonal BD; the two triangles ABD, BDC, have the three sides of the one equal to the three THEOREM. Fig. 44. as † A re-entering angle is one whose vertex is directed inward, Fig. 43. CDE (fig. 43), while a saliant angle has its vertex directed outward as ABC |