28 Book I. the other fides fhall be equal, each to each, viz. AB to DE, and AC to DF; and the third angle BAC to the third angle EDF. a 4. 1. a For, if AB be not equal to DE, one of them must be the greater. 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 fides GB, BC are equal to the two DE, FF, each to each; and the angle GBC is equal to the angle DEF; therefore the bafe GC is equal to the bafe DF, and the triangle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal fides are oppofite; therefore the angle GCB is equal to the angle DFE; but DFE is, by the hypothefis, equal to the angle BCA; wherefore alfo the angle BCG is equal to the angle BCA, the lefs to the greater, which is impoffible; 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; the base therefore AC is equal to the bafe DF, and the third angle BAC to the third angle EDF. a Next, let the fides which are oppofite to equal angles in each triangle be equal to one another, viz. AB to DE; likewife in this cafe, the other fides fhall 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 join AH; and because BH is 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 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 fides are oppofite; therefore the angle BHA is equal to the angle a EFD; EFD; but EFD is equal to the angle BCA; therefore alfo Book I. 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 oppofite angle BCA; which is impoffible b; wherefore b 16. 1. BC is not unequal to EF, that is, it is equal to it; and AB is equal to D E; 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. Therefore, if two triangles, &c. Q. E. D. IF PROP. XXVII. THEOR. [F a ftraight line falling upon two other ftraight lines makes the alternate angles equal to one another, these two straight lines are parallel. Let the ftraight line EF, which falls upon the two ftraight 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 a than the a 16. 1. interior and oppofite manner it may be de monstrated that they do not meet towards A, C; but those ftraight lines which meet neither way, though produced ever fo far, are parallel b to one another. AB therefore is parallel b 30. Def. to CD. Wherefore, if a straight line, &c. Q. E. D. PROP. Book I. 2 15. 1. b 27. 1. See N. PROP. XXVIII. THEOR. TF a ftraight line falling upon two other ftraight lines makes the exterior angle equal to the interior and oppofite upon the fame fide of the line; or makes the interior angles upon the fame fide together equal to two right angles; the two ftraight lines are parallel to one another. Let the ftraight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and oppofite angle GHD upon the same fide; on the fame fide BGH, A Because the angle EGB C is equal to the angle GHD, and the angle EGB equal a E d angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel b to CD. Again, because the angles BGH, GHD are equal to two right angles; and that AGH, BGH, are alfo equal to two right angles; the angles AGH, BGH are equal to the angles BGH, GHD: Take away the common angle BGH; there- ' fore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. PROP. XXIX. THEOR. Fa ftraight line fall upon two parallel ftraight lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and oppofite upon the fame fide; and likewife the two interior angles upon the fame fide together equal to two right angles, Let Let the straight line EF fall upon the parallel straight lines Book I. AB, CD; the alternate angles AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and oppofite, upon the fame fide, GHD; and the two interior angles BGH, GHD upon the fame fide are together equal to two right angles. E G L a 27. I, 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; but AB is also parallel to CD; there- A fore two ftraight lines. are drawn through the K fame point G, parallel to CD, and yet not coinciding with one C D H F another, which is im poffible. The angles AGH, GHD therefore are not unequal, that is, they are equal to one another. Now, the angle EGB is equal to AGH; and AGH is proved to be equal to GHD; there- c 15. 1. fore EGB is likewise 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 to d 13. I, two right angles; therefore alfo BGH, GHD are equal to two right angles. Wherefore, if a ftraight line, &c. Q. E. D. d COR. If two lines KL and CD make, with EF, the two angles KGH, GHC together lefs than two right angles, KG and CH will meet on the fide 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 fide 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 fide of EF; for the angles LGH, GHD would then be two angles of a triangle, and lefs than two right angles; but this is impoffible; for the four angles KGH, HGL, CHG, GHD are together equal to four right angles d, of which the two, KGH, CHG are by fuppofition lefs than two right angles; therefore the other two, HGL, GHD are greater than two right angles. Therefore, fince KL and CD are not parallel, and do not meet towards Land D, they will meet if produced towards K and C. PROP. 7 Book I. STR PROP. XXX. THEOR. a TRAIGHT lines which are parallel to the fame ftraight line are parallel to one another. Let AB, CD be each of them parallel to EF; AB is alfo parallel to CD. Let the ftraight line GHK cut AB, EF, CD; and because GHK cuts the parallel ftraight lines AB, EF, the angle was fhewn that the angle C is equal to GKD; and they are alternate angles; therefore 27.1. AB is parallel to CD. Wherefore ftraight lines, &c. Q E. D. PROP. XXXI. PROB. O draw a ftraight line through a given point T parallel to a given ftraight line. Let A be the given point, and BC the given straight line; it is required to draw a A ftraight line through the E point A, parallel to the ftraight line BC. In BC take any point B D, and join AD; and at a 23. 1. the point A, in the straight line AD, make the angle DAE equal to the angle ADC; and produce the straight line EA to F. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal 27. 1. to one another, EF is parallel to BC. Therefore the straight line |