equal to DE, and BC to EF, the two sides GB, BC, are equal Book I. 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 a 4. 1. 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 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; the base therefore AC is equal to the base DF, and the third angle BAC to the third angle EDF. Next, let the sides D B HC E to FF; 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 bafe AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles shall be 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 oppolite angle BCA ; which is impossible b; wherefore BC is not unequal to EF, that is, 6 16.1. 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. Therefore, if two triangles, &c. Q. E. D. PRO P. IF Book T. PRO P. XXVII. TH E O R. makes 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 is parallel to CD. For, if it be not parallel, AB and CD being produced fhall 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 than the interior and opposite angle EFG; but it is also equal to it, which is impossible; there A E B F D those straight lines which meet neither way, though 8 35. Def. produced ever so far, are parallel b to one another. AB therefore is parallel to CD. Wherefore, if a straight line, &c. Q. E D. 16. I. PRO P. XXVIII. THE OR. makes the exter or angle equal to the interior and opposite upon the same fide of the line; or inakes the interior angles upon the same side together equal to two right angles; the two straight lines fhail be parallel to one anottier. Let the straight line EF, which E -D H F angle angle EGB equal to the angle AGH, the angle AGH is equal Book T. to the angle GHD; and they are the alternate angles; therefore AB is parallel to CD. Again, because the angles BGH, GHD a 15. 1. are equal to two right angles, and that AGH, BGH are also • 25. 1. equal to two right angles; the angles AGH, BGH are equal : 13.1. c By Hyp, 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. Wherefore, if a straight line, &c. Q. E. D. PRO P. XXIX. THE O R. IF a straight fire falls upon two parallel straight lines, it See the makes the alternate angles equal to one another; and this propothe exterior angle equal to the interior and opposite upon the fane fide, and likwise the two interior angles upon the fame fide together equal to two right angles. Gition. Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and oppofte, upon the same side, E GHD; and the two interior angles BGH, GHD upon the fame fide are together equal to two right angles. A B For, if AGH be not equal to GHD, one of them must be greater than the other ; let AGH be the greater; and С H D because the angle AGH is greater F than the angle GHD, add to each of them the angle BGH; therefore the angles AGH, BGH are than the angles BGH, GHD; but the angles AGH, BGH are equal to two right angles; therefore the angles a 13. I. BGH, GHD are less than two right angles; but those straight lines which, with another straight line falling upon them, make the interior angles on the same fide less than two right angles, do meet * together if continually produced; therefore the 19. ax. Itraight lines AB, CD, if produced far enough, shall meet; but See the they never meet, since they are parallel by the hypothesis ; notes on therefore the angle AGH is not unequal to the angle GHD, this propes that is, it is equal to it; but the angle AGH is equal to the 615.8 angle EGB; therefore likewise EGB is equal to GHD; add to C 2 cach Book I. each of thele the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are e, 13. I. qual o to two right angles; therefore also BGH, GHD are e qual to two right angles. Wherefore, if a straight line, &c. Q. E. D. PROP. XXX. THE OR. ST a 29. 1. Traight lines which are parallel to the same straight line are parallel to one another. Let AB, CD be each of them parallel to EF; AB is also parallel to CD. Let the straight line GHK cut AB, EF, CD; and because G -B Н. -F K D b 27. 1, To draw a straight line through a given point paral lel a Let A be the given point, and BC the given straight line; it is required to draw a straight line Α. F In BC take any point D, and join AD; and at the point A in the DAE equal to the angle ADC; and Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to b 27, I. one another, EF is parallel to BC. Therefore the straight line EAF EAF is drawn through the given point A parallel to the given Book I. straight line BC. Which was to be done. IF a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB are together equal to two right angles. Through the point C draw CE parallel to the straight a 31. 8. line AB; and because AB is А parallel to CE and AC meets E them, the alternate angles BAC, ACE are equal b. A b 29. I. gain, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD B O D is equal to the interior and oppolite angle ABC; but the angle ACE was shown to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these equals add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB ; but the angles ACD, ACB are equal to two right angles; therefore also the c 13. 1. angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D. COR. I. All the interior angles of any rectilineal figure, together D with four right angles, are equal to twice as many right angles as E the figure has fides. For any rectilineal figure ABCDL can be divided into as F many triangles as the figure has fides, by drawing straight lines from a point F within the figure A B |