20. . of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle: BD and DC shall be less than the other two sides BA, AC of the triangle, but shall contain an angle BDC greater than the angle BAC. Produce BD to E: and because two sides of a triangle * are greater than the third side, the two sides BA, AE of the triangle ABE, E are greater than BE: to each of these add EC; therefore the sides BA, AC are greatert + 4 Ax. в. than BE, EC. Again, because the two sides CE, ED of the triangle CED, are greater + than CD, add DB † 20. 1. to each of these ; therefore the sides CE, EB are greater of † 4 Ax. than CD, DB: but it has been shewn that BA, AC are greater than BE, EC ; much more then are BA, AC greater than BD, DC. Again, because the exterior angle of a triangle * is greater • 16. 1. than the interior and opposite angle, the exterior angle BDC of the triangle CDE, is greater than CED: for the same reason, the exterior angle CEB of the triangle ABE, is greater than BAC: and it has been demonstrated, that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q. E. D. PROPOSITION XXII. PROB.—To make a triangle of which the sides shall be equal See N. to three given straight lines, but any two whatever of these must be greater than the third *. * 20.1. Let A, B, C be the three given straight lines, of which any two whatever are greater than the third; viz. A and B greater than C; 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 K at the point D, but unlimited towards E, and make * DF equal to A, FG D ME * 3. 1. equal to B, and GH equal to C: from the centre F, at the distance FD, describe * the circle DKL ; and from the * 3 Post, centre G, at the distance GH, describe * another circle HLK ; and join KF, KG: the triangle KFG shall have its sides equal to the three straight lines A, B, C. 3 Post. * 15 Def. + Constr. + 1 Ax. * 15 Def. + Constr. Because the point F is the centre of the circle DKL, FD is equal * to FK; but FD is equal + to the straight line A ; therefore FK is equal † to A : again, because G is the centre of the circle LKH, GH is equal * to GK; but GH is equal to C; therefore also GK is equal to C: and FG is equal + to B; therefore the three straight lines KF, FG, GK are 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. Which was to be done. PROPOSITION XXIII. PROB. — At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. ДД EFA * 22. 1. Let AB be the given straight line, and A the given point in D B Because DC, CE are equal to FA, AG, each to each, and the base DE to the base FG, the angle DCE is equal * to the angle FAG. Therefore, at the given point A, in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done. PROPOSITION XXIV. See N. THEOR.-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 ; viz. AB equal 23. I. to DE, and AC to DF, but the angle BAC greater than the angle EDF: the base BC shall be greater than the base EF. Of the two sides DE, DF, let.DE be the side which is not greater than the other and at the point D, in the straight line DE, make * the angle EDG equal to the angle BAC ; and make DG equal * to AC or DE, and jóín EG, GF.' Because AB is equal + to DE, and AC'+ to DG, the two sides | Hyp. + Constr. BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal t to the angle + Constr. EDG; therefore the base BC is equal * to the base EG. And be • 4. 1. cause DG is equal to DF, the angle DFG is equal * to the angle DGF; G: 5. 1. but the angle DGF is greater + than the angle EGF; therefore the angle DFG is greater than EGF; therefore, much more is the angle EFG, greater than the angle EGF: and because the angle EFG of the triangle EFG, is greater than its angle EGF, and that the greater * angle is * 19. 1. subtended by the greater side, therefore the side EG is greater than the side EF: but EG was proved to be equal to BC; therefore BC is greater than EF. Therefore, if two triangles, &c. QoE. DE PROPOSITION XXV. Theor.-If two triangles have two sides of the one, equal to See N. two sides of the other, each to each, but the base of the one, greater than the base of the other, the angle contained by the sides of that 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 ; viz. AB equal to DE, and AC to DF, but the base BC greater than the base EF: the A D angle BAC shall be greater than the angle EDF. For if it be not greater, it must either be equal to it, or less than it: B but the angle BAC is not equal to the angle EDF, because then the base BC would be equal * to EF; but it is t not ; 4.1. therefore the angle BAC is not equal to the angle EDF: neither + Hyp. E 24. I. + Hyp. is it less, because then, the base BC would be less * than the base EF; but it is t not; therefore the angle BAC is not less than the angle EDF: and it was shewn, that it is not equal to it; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E. D. PROPOSITION XXVI. G E TheoR.-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, or the sides opposite to equal angles in each, then shall the other sides be equal, each to each, and also the third angle of the one, to the third angle of the other. Let ABC, DEF be two triangles, which have the angles D B For if AB be not equal to DE, one of them must be greater than the other: let AB be the greater of the two, and make BG equal + to DE, and join GC: therefore, because BG is equal to DE, and BC + to EF, the two sides GB, BC are equal to the two DE, EF, each tò each ; and the angle GBC is equal + to the angle DEF; therefore the base GC is equal to the base DF, 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 DFE is, by the hypothesis, equal to the angle BCA ; wherefore also the angle BCG is equal + to the angle BCA, the less to the greater ; which is impossible; therefore AB is not unequal to DE, that is, it is equal to it: and BC if equal + to EF: therefore the two AB, BC are equal to the two DE, EF, each to each ; and the angle ABC is equal + to the angle DEF; therefore the base AC is ual * to the base DF, and the third angle BAC to the third angle EDF. + 3. 1. + Hyp: + Hyp. * 4.1. +1 Ax. + Hyp. + Hyp. * 4. 1. Next, let the sides which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE ; likewise in this case, the other sides shall be equal, AC to DF, and BC to EF; and also the third D angle BAC to the third angle EDF. For if BC be not equal to EF, let BC be the greater of them, and make BH equal t to EF, and join AH; and † 8. 1. B HC E because BH is equal to EF, and AB to + DE, the two AB, BH are equal to the two DE, EF, + Hyp. each to each ; and they contain equal + angles; therefore + Hyp. the base AH is equal + to the base DF, and the triangle ABH † 4. 1. 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; but EFD is equal + to the angle BCA; therefore also the angle + Hyp. BHA is equal to the angle BCA; that is, the exterior angle † 1 Ax. BHA of the triangle AHC, is equal to its interior and opposite angle BCA ; which is impossible *: wherefore BC is not unequal to EF, that is, it is equal to it: and AB is equal + to † Hyp. DE; therefore the two AB, BC are equal to the two DE, EF, each to each ; and they contain + equal angles : wherefore the + Hyp. base AC is equal † to the base DF, and the third angle BAC, † 4. 1. to the third angle EDF. Therefore, if two triangles, &c. Q. E. D. . 16. 1. PROPOSITION XXVII. THEOR.-If a straight line, falling upon two other straight lines, make the alternate angles equal to one another, 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 : AB shall be parallel to CD. For if it be not parallel, AB and CD, being produced, will meet either towards B, D, or towards A, C: let them be pro- C7 duced and meet towards B, D, in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater * than the interior and opposite angle EFG ; but it is * 16. 1. also equal + to it; which is impossible ; therefore, AB and CD, † Hyr. being produced, do not meet towards B, D. In like manner |