the centre F, at the distance FD, describe (3. Post.) the circle DKL, and from the centre G, at the distance GH, describe (3. Post.) another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines, A, B, C. Because the point F is the centre of the circle DKL, FD is equal (11. Def.) 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 (11. Def.) 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. SCHOLIUM. If one of the sides were greater than the sum of the other two, the arcs would not intersect each other: but the solution will always be possible, when the sum of two sides, any how taken (20. 1.) is greater than the third. 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. Take in CD, CE any points D, E, and join DE; and make (22. 1.) the triangle AFG, the sides of which shall be equal to the three straight lines, CD, DE, CE, so that CD be equal to AF, CE to AG, and DE to FG; and because E A G F DC, CE are equal to FA, AG, D each to each, and the base DE to the base FG; the angle DCE is equal (8. 1.) to the angle FAG. B Therefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. 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 the one greater than the angle contained by the two sides 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 to DE, and AC to DF; but the angle BAC greater than the angle EDF; the base BC is also 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 (23. 1.) the angle EDG equal to the angle BAC: and make DG equal (3. 1.) to AC or DF, and join EG, GF. Because AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal to the angle EDG, therefore the base BC is equal (4.1.) to the base EG; and because DG is equal to DF, the angle DFG is equal (5. 1.) to the angle DGF; but the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF; and because AA G F the angle EFG of the triangle EFG is greater than its angle EGF, and because the greater (19. 1.) side is opposite to the greater angle, the side EG is greater than the side EF; but EG is equal to BC; and therefore also BC is greater than EF. PROP. XXV. THEOR. If two triangles have two sides of the one equal to 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 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 let the base CB be greater than the base EF, the angle BAC is likewise greater than the angle EDF. For, if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle EDF, because then the base BC would be equal (4. 1.) to EF; but it is not; therefore the angle BAC is not equal to the angle EDF; neither is it less; because then the base BC would be less (24. 1.) than the bas EF; but it is 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. B AA 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 the equal angles in each; then shall the other side 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 ABC, BCA equal to the angles DEF, EFD, viz. ABC to DEF, and BCA to EFD, also one side equal to one side; and first, let those sides be equal which are adjacent to the angles that are equal in the two triangles, viz. BC to EF; the other sides shall be equal, each to each, viz. D AA AB to DE, and AC to DF; and B the third angle BAC to the third angle EDF. C E F Let For, if AB be not equal to DE, one of them must be the greater. 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 to each; and the angle GBC is equal to the angle DEF; therefore the base GC is equal (4. 1.) 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 is 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 equal (4. 1.) to the base DF, and the angle BAC to the angle EDF. 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 angle BAC to the third EDF. For, if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and B join AH; and because BH is HCE equal to EF, and AB to DE; the two AB, BH are equal to the two DE, EF, each to each; and they contain equal angles; therefore (4. 1.) the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles are equal, 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 BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA, which is impossible (16. 1.); wherefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE; therefore the two, AB, BC are equal to the two DE, EF, each to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. PROP. XXVII. THEOR. If a straight line falling upon two other straight lines makes the alternate angles equal to one another, these two straight lines are 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 is parallel to CD. For, if it be not parallel, AB and CD being produced shall meet either towards B, D, or towards A, C; let them be produced and meet towards B, D in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater (16. 1.) than the interior and opposite angle EFG; but it is also equal to it, which is impossible therefore, AB and CD being produced, do not meet towards B, D. In like manner it may be demonstrated that they do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel (30. Def.) A E B G C F D to one another. AB therefore is parallel to CD. PROP. XXVIII. THEOR. 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 are 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 GHD, the interior and opposite angle upon the same side; or let it make the interior angles on the same side BGH, GHD together equal to two right angles; AB is parallel to CD. Because the angle EGB is equal to the angle GHD, and also (15. 1.) to the A E G B H D F angle AGH, the angle AGH is equal to the angle GHD; and they are th alternate angles; therefore AB is parallel (27. 1.) to CD. Again, because the angles BGH, GHD are equal (hyp.)to two right angles, and AGH, BGH, are also equal (13. 1.) to two right angles, the angles AGH, BGH are equal to the angles BGH, GHD: Take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD. COR. Hence, when two straight lines are perpendicular to a third line, they will be parallel to each other. PROP. XXIX. THEOR. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles. Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHI) are equal to one another; and the exterior angle EGB is equal to the interior and opposite, upon the same side, GHD; and the two interior angles BGH, GHD upon the same side are together equal to two right angles. E A G L -B For if AGH be not equal to GHD, let KG be drawn making the angle KGH equal to GHD, and produce KG to L; then KL will be parallel to CD (27.1.); but AB is also parallel to CD; therefore two straight lines are drawn through the same point G, parallel to CD, and yet not coinciding with one another, which is impossible (11. Ax.) The angles AGH, GHD therefore are not unequal, that is, they are equal to one another. Now, the angle EGB is equal to AGH (15. 1.); and AGH is proved to be equal to GHD; therefore EGB is like K C H -D wise equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal (13. 1.) to two right angles; therefore also BGH, GHD are equal to two right angles. COR. 1. If two lines KL and CD make, with EF, the two angles KGH, GHC together less than two right angles, KG and CH will meet on the side of EF on which the two angles are that are less than two right angles. For, if not, KL and CD are either parallel, or they meet on the other side of EF; but they are not parallel; for the angles KGH, GHC would then be equal to two right angles. Neither do they meet on the other side of EF; for the angles LGH, GHD would then be two angles of a triangle, and less than two right angles; but this is impossible; for the four angles KGH, HGL, CHG, GHD are together equal to four right angles (13. 1.) of which the two, KGH, CHG, are by supposition less than 2 |