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Fig. 37

THEOREM.

71. If two straight lines AB, CD (fig. 37), make with a third EF, two interior angles, on the same side, the sum of which is greater or less than two right angles, the lines AB, CD, produced sufficiently far, will meet.

Demonstration. 1. Let the sum BEF + EFD be less than two right angles; draw FG so as to make the angle EFG equal to AEF; we shall have the sum BEF + EFG equal to the sum BEF+AEF, and consequently equal to two right angles; and, since BEF+ EFD is less than two right angles, the straight line DF will be comprehended in the angle EFG.

Through the point F draw an oblique line FM, meeting AB in M; the angle AMF will be equal to GFM, since, by adding to each the same quantity EFM+ FEM, the two sums are each equal to two right angles. Take now MN- FM, and join FN; the exterior angle AMF, of the triangle FMN, is equal to the sum of the two opposite interior angles MFN, MNF (63), and these last are equal to each other, since they are opposite to the equal sides MN, FM; consequently the angle AMF, or its equal MFG, is double of MFN; therefore the straight line FN divides into two equal parts the angle GFM, and meets the line AB in a point N situated at a distance MN equal to FM.

It follows from the same demonstration, that if we take NP = FN, we determine, upon the line AB, the point P of the straight line FP, which makes the angle GFP equal to half the angle GFN, or one fourth of the angle GFM.

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We are able, therefore, in this manner, to take successively the half, the fourth, the eighth, &c., of the angle GFM, and the lines which form these divisions meet the line AB in points more and more distant, but easily determined, since MN FM, NPFN, PQ=PF, &c. Indeed, it will be remarked, that each successive distance of the points of intersection from the fixed point F, is not exactly double the distance of the preceding point of intersection; since FN, for example, is less than FM + MN, or 2 FM; we have, in like manner, FP <2 FN, FQ< 2 FP, &c.

But, by continuing to subdivide the angle GFM, in this manner,

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we shall soon arrive at an angle GFZ less than the given angle GFD; and it will nevertheless be true that FZ produced will meet AB in a determinate point; therefore, for a still stronger reason, the straight line FD, comprehended in the angle EFZ, will meet AB.

2. Let us suppose that the sum of the interior angles AEF+ CFE is greater than two right angles; if we produce AE toward B, and CF toward D, the sum of the four angles AEF, BEF, CFE, EFD, will be equal to four right angles; accordingly, if from this sum we subtract AEF + CFE, which is greater than two right angles, there will remain the sum BEF+ EFD, less than two right angles. Therefore, according to the first case, the lines EB, FD, produced sufficiently far, must meet. 72. Corollary. Through any given point F only one parallel can be drawn to the given line AB; for, having drawn FE at pleasure, there can be no other line, except FG, which shall make the sum of the two angles BEF + EFG equal to two right angles; every other straight line FD would make the sum of the two angles BEF+ EFD, either less or greater than two right angles; and would consequently meet the line AB.

THEOREM.

73. If two parallel straight lines AB, CD (fig. 38), meet a Fig. 38 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 (71).

74. Corollary 1. 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.

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75. Corollary II. Since the sum AGH+ GHC is equal to two right angles, and the sum GHD+ GHC is also equal to two right 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

Fig. 39.

Fig. 40.

the four acute angles to one of the four obtuse angles, the sum will always be equal to two right angles.

76. Scholium. The angles of which we have been speaking, compared, two and two, take different names. We have already called the angles AGH, 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.

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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 we have also AGH+ GHC equal to two right angles; therefore the lines AG, CH, are parallel (70).

THEOREM.

77. Two lines AB, CD (fig. 39), which are parallel to a third 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 (74); 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 parallel (69).

THEOREM.

78. Two parallel lines are throughout at the same distance from each other.

Demonstration. The two parallels AB, CD (fig. 40), being 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 (73); moreover these straight lines will be equal to each other.

For, by drawing HE, the angles HEF, EHG, considered with reference to the parallels AB, CD, being alternate-internal angles (76), 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 EHF, HEG, considered with reference to the parallels GE, FH, being alternate-internal angles, are equal. The two triangles, then, EHF, HEG, 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 distance of the same parallels at the point F.

THEOREM.

79. If two angles, BAC, DEF (fig. 41), have their sides par- Fig. 41, allel, each to each, and directed the same way, these two angles will be equal.

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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 (76); the angle DGC is equal to BAC, because DG is parallel to AB; therefore the angle DEF is equal to BAC.

80. 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.

THEOREM.

81. The opposite sides of a parallelogram are equal, and the opposite angles also are equal.

Demonstration. Draw the diagonal BD (fig. 44); the two Fig. 44 triangles ADB, DBC, have the side BD common; moreover, on account of the parallels AD, BC, the angle ADB = DBC (76), and on account of the parallels AB, CD, the angle ABD=BDC; therefore the two triangles ADB, DBC, are

Fig. 44.

Fig. 44.

Fig. 45.

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 AC, 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.

82. Corollary. Hence two parallels AB, CD, comprehended between two other parallels AD, BC, are equal.

THEOREM.

83. 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 sides of the other, each to each; they are therefore equal, and the angle ADB, opposite to the side AB, is equal to the angle DBC, opposite to the side CD; consequently the side AD is parallel to BC (76). For a similar reason, AB is parallel to CD; therefore the quadrilateral ABCD is a parallelogram.

THEOREM.

84. If two opposite sides AB, CD (fig. 44), of a quadrilateral are equal and parallel, the two other sides will also be equal and parallel, and the figure ABCD will be a parallelogram.

Demonstration. Let the diagonal BD be drawn; since AB is parallel to CD, the alternate angles ABD, BDC, are equal (76). Besides, the side ABCD, and the side DB is common; therefore the triangle ABD is equal to the triangle DBC (36), and the side AD = BC, the angle ADB = DBC, and consequently AD is parallel to BC; therefore the figure ABCD is a parallelogram,

THEOREM.

85. The two diagonals AC, DB (fig. 45), of a parallelogram mutually bisect each other.

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