The proposition will be evidently true if it can be proved that BC is equal to EF (theorem 10). Let us suppose, if it be possible, that these sides are anequal, and that BC is the greater. Take BG equal to EF, and join AG. The triangles ABG and DEF, having AB equal to DE, and BG equal to EF, by hypothesis, and also having the angle ABG equal to DEF, they will be equal THEOREM 17. 68. Two straight lines perpendicular to a third are parallel. For, if the straight lines AC, BD, be not parallel, they —=^ will meet on one side or the other of the line AB; let them meet in O; then AC and OB are both perpendi- —g Jj cular to AB, from the same point O; which is impossible (theorem 13). Theorem 18. 69. If two straight lines, AC and BD, make, with a third, AB, the sum of the two interior angles CAB, ABD, equal to two right angles, these two straight lines are parallel. From G, the middle of AB, draw EGF, perpendicular -^to AC: then, since the sum of the angles ABD, ABF, is equal to two right angles the sum of the two angles ABD, BAC, is also equal to two right angles; therefore the two angles ABD, ABF, are together equal to the sum of the two angles ABD, BAC; and, taking away the common angle ABD, there remains the angle ABF = BAC ; that is, GBF equal to GAE. But the angles BGF and AGE are also equal (theorem 3); and, since BG is equal to GA, therefore the triangles BGF and AGE, having a side and two adjacent angles of the one equal to a side and two adjacent angles of the other, are equal {theorem 6),
and the angle BFG is equal to AEG; but AEG is, by construction, a right angle; therefore, BFG is also a right angle; and, since GEC is a right angle, the straight lines EC and FD are perpendicular to EF, and are, therefore, parallel to each other THEOREM 19. 70. If two straight lines, AC, BD, make with a third, HK, the alternate angles, AHK and HKD,#equal, the two lines are parallel. For, adding KHC to each of the angles AHK, HKD, the . sum of the angles AHK, KHC, is equal to the sum of the angles HKD, KHC; but the angles AHK, KHC, are to- B_ gether equal to two right angles; therefore, also, the angles /fi HKD, KHC, are also equal to two right angles; and, consequently, AC is parallel to BD (theorem 18). THEOREM 20. 71. If two straight lines, AC, BD, are cut by a third, FG, so as to make the exterior angle, FHC, equal to the interior and opposite angle, HGD, on the same side, the two lines are parallel. For, since the angle FHC is equal to the angle AHG, and since, when AC is parallel to BD, the angle AHG is equal to min (theorem 19), therefore the angle FHC is equal to HGD. Theorem 21. 72. If a straight line, EF, meet two parallel straight lines, AC, BD, the sum of the inward angles CEF, EFD, on the same side, will be equal to two right angles. For, if not, suppose EG to be drawn through E, so that M EL C the sum of the angles GEF and EFD may be two right x^ff angles; then EG will be parallel to BD (theorem 18); and g -r —5 thus, through the same point E, two straight lines, EG, EC, are drawn, each parallel to BD; which is impossible (ax. 11, p. 15); therefore no straight line that does not coincide with AC, is parallel to BD; wherefore the straight line AC is parallel to BD. 73. Corollary.—If a straight line is perpendicular to one of two parallel straight lines, it is also perpendicular to the other. THEOREM 22. 74. If a straight line, HK, meet two parallel straight lines, AC, BD, the alternate angles, AHK, HKD, shall be equal. For, the sum of the angles CHK, HKD, is equal to two be equal to AHK. Theorem 23. 75. If a straight line FG, cut two parallel straight lines, AC, BD, the exterior angle, FHC, is equal to the interior and opposite angle HKD. For, since the angle FHC is equal to the angle AHK A ^/F c {theorem 3), and the angle AHK equal to the angle HKD; therefore the angle FHC is equal to the angle HKD. Theorem 76. If a straight line, EF, meet two other straight lines, EG, FD, and make the two interior angles, EFD, FEG, on the same side, less than two right angles, the lines EG, FD, meet, if produced, on the side of EF, on which the angles are less than two right angles. For, if they do not meet on that side, they are either parallel, or else they meet on the other side. Now they cannot be parallel, for then the two interior angles would be equal to two right angles, instead of being less. Again, to show that they cannot meet on the other side, suppose EA to be parallel to DFB; then, because the sum "^ of the angles EFD, FEG, is, by hypothesis less than two right angles, that is, less than the sum of the two angles, ^ FEK, FEG The truth of this proposition is assumed as an axiom in the Elements of Euclid, and made the foundation of parallel lines. Theorem M. 77. Two straight lines, AB, CD, parallel to a third, EF, are parallel to one another. Draw the straight line PQR, perpendicular to EF. Because AB is parallel to EF, the line PR shall be perpendicular to AB; and, because CD is parallel to EF, the line PR is also perpendicular to CD: therefore AB and CD are perpendicular to the same straight line PQ; hence they are parallel {theorem 17). Theorem 26. 78. Two parallel straight lines are every where equally distant. Let AB, CD, be two parallel straight lines. From any c? 9"" points, E ahd F, in one of them, suppose perpendiculars EG, s^ FH, to be drawn; these, when produced, will meet the Ap as others at right angles, in H and G. Join FG; then, because FH and EG are both perpendicular to AB, they are parallel THEOREM 27. 79. In any triangle, if one of the sides be produced, the exterior angle is equal to both the interior and opposite angles; and the three interior angles are equal to two right angles. Let ABC be a triangle; produce any one of its sides, AC B towards D; and, from the point A, let AE be drawn, parallel to BC; and, because of the parallels CB and AE, and the <* A » angle EAD = C, and the angle EAB - B Again, because the angle BAD is equal to the sum of the angles B and C, add to each the angle BAC, and the sum of the two angles BAC, BAD, will be equal to the sum of the three angles BAC, B, C, or the three angles of the triangle; but the sum of the two angles BAC, BAD, is equal to two right angles 80. Corollary 1.—If two angles of one triangle be equal to two angles of another triangle, each to each, the third angle of the one shall be equal to the third angle of the other, and the triangles shall be equi-angular. 81. Corollary 2.—A triangle can have only one right angle. 82. Corollary 3.—In any right-angled triangle the sum of the two acute angles is equal to a right angle. 83. Corollary 4.—In an equilateral triangle, each of the angles is onethird of two right angles. |