PROPOSITION XXII. PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third. Let A, B, C be the three given straight lines, of which any two whatever are greater than the third, (1. 20.) A and C greater than B; and B and C greater than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but unlimited towards E, make DF equal to A, FG equal to B, and GH equal to C; (1. 3.) from the centre F, at the distance FD, describe the circle DKL; (post. 3.) and from the centre G, at the distance GH, describe the circle HLK; and join KF, KG. Then the triangle KFG shall have its sides equal to the three straight lines A, B, C. Because the point F is the centre of the circle DKL, but FD is equal to the straight line A; Again, because G is the centre of the circle HLK; but GH is equal to C; therefore also GK is equal to C; and FG is equal to B; (ax. 1.) therefore the three straight lines KF, FG, GK, are respectively equal to the three, A, B, C : and therefore the triangle KFG has its three sides KF, FG, GK, equal to the three given straight lines A, B, C. Q. E. F. PROPOSITION XXIII. PROBLEM. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle. It is required to make an angle at the given point A in the given straight line AB, that shall be equal to the given rectilineal angle DCE. In CD, CE, take any points D, E, and join DE; make the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC, so that AF be equal to CD, AG to CE, and FG to DE. (1.22.) Then the angle FAG shall be equal to the angle DCE. and the base FG is equal to the base DE; therefore the angle FAG is equal to the angle DCE. (1.8.) Wherefore at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Q.E.F. PROPOSITION XXIV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other; the base of that which has the greater angle, shall be greater than the base of the other. Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, namely AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF. Then the base BC shall be greater than the base EF. F Of the two sides DE, DF, let DE be not greater than DF, make the angle EDG equal to the angle PAC; (1. 23.) make DG equal to DF or AC, (1. 3.) and join EG, GF. Then, because DE is equal to AB, and DG to AC, the two sides DE, DG are equal to the two AB, AC, each to each, and the angle EDG is equal to the angle BAC; therefore the base EG is equal to the base BC. (1.4.) And because DG is equal to DF in the triangle DFG, therefore the angle DFG is equal to the angle DGF; (1. 5.) but the angle DGF is greater than the angle EGF; (ax. 9.) therefore the angle DFG is also greater than the angle EGF; much more therefore is the angle EFG greater than the angle EGF. And because in the triangle EFG, the angle EFG is greater than the angle EGF, and that the greater angle is subtended by the greater side; (1. 19.) therefore the side EG is greater than the side EF; but EG was proved equal to BC; therefore BC is greater than EF. Wherefore if two triangles, &c. Q.E.D. PROPOSITION XXV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other; the angle contained by the sides of the one which has the greater base, shall be greater than the angle contained by the sides, equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, namely, AB equal to DE, and AC to DF; but the base BC greater than the base EF. Then the angle BAC shall be greater than the angle EDF. For, if the angle BAC be not greater than the angle EDF, it must either be equal to it, or less than it. If the angle BAC were equal to the angle EDF, then the base BC would be equal to the base EF; (1. 4.) therefore the angle BAC is not equal to the angle EDF. therefore the angle BAC is not less than the angle EDF; and it has been shewn, that the angle BAC is not equal to the angle EDF; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E. D. PROPOSITION XXVI. THEOREM. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal angles in each, or the sides opposite to them; then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, each to each, namely ABC to DEF, and BCA to EFD; also one side equal to one side. First, let those sides be equal which are adjacent to the angles that are equal in the two triangles, namely BC to EF. Then the other sides shall be equal, each to each, namely AB to DE, and AC to DF, and the third angle BAC to the third angle EDF. For, if AB be not equal to DE, one of them must be greater than the other. Let AB be greater than DE, make BG equal to ED, (1. 3.) and join GC. because GB is equal to DE, and BC to EF, (hyp.) the two sides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is equal to the angle DEF; therefore the base GC is equal to the base DF, (1. 4.) and the triangle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle GCB is equal to the angle DFE; but the angle DFE is, by the hypothesis, equal to the angle ACB; wherefore also the angle GCB is equal to the angle ACB; (ax. 1.) the less angle equal to the greater, which is impossible; therefore AB is not unequal to DE, that is, AB is equal to DE. Hence, in the triangles ABC, DEF; because AB is equal to DE, and BC to EF, (hyp.) and the angle ABC is equal to the angle DEF; (hyp.) therefore the base AC is equal to the base DF, (1. 4.) and the third angle BAC to the third angle EDF. Secondly, let the sides which are opposite to the equal angles in each triangle be equal to one another, namely, AB equal to DE. Then in this case likewise the other sides shall be equal, AC to DF, and BC to EF, and also the third angle BAC to the third angle EDF. For if BC be not equal to EF, one of them must be greater than the other. Let BC be greater than EF; make BH equal to EF, (1.3.) and join AH. Then in the two triangles ABH, DEF, because AB is equal DE, and BH to EF, and the triangle ABH to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle BHA is equal to the angle EFD; equal to its interior and opposite angle BCA; which is impossible; (1.16.) wherefore BC is not unequal to EF, that is, BC is equal to EF. Hence, in the triangles ABC, DEF; because AB is equal to DE, and BC to EF, (hyp.) and the included angle ABC is equal to the included angle DEF; (hyp.) therefore the base AC is equal to the base DF, (1.4.) and the third angle BAC to the third angle EDF. Wherefore if two triangles, &c. PROPOSITION XXVII. Q.E.D. THEOREM. If a straight line, falling on two other straight lines, make the allernate angles equal to each other; these two straight lines shall be parallel. Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another. Then AB shall be parallel to CD. For, if AB be not parallel to CD, AB and CD being produced will meet either towards A and C, or towards B and D. Let AB, CD be produced and meet towards B and D, in the point G. Then GEF is a triangle, and its exterior angle AEF is greater than the interior and opposite angle EFG; (1.16.) but the angle AEF is equal to the angle EFG; (hyp.) therefore the angle AEF is greater than and equal to the angle EFG; which is impossible. Therefore AB, CD being produced do not meet towards B, D. In like manner, it may be demonstrated, that they do not meet when produced towards A, C. But those straight lines in the same plane which meet neither way, though produced ever so far, are parallel to one another; (def. 35.) therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q.E.D. PROPOSITION XXVIII. THEOREM. If a straight line falling upon two other straight lines, makes the exterior angle equal to the interior and opposite upon the same side of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another. Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD upon the same side; or make the two interior angles BGH, GHD on the same side together equal to two right angles. Then AB shall be parallel to CD. E B H |