Book I. 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 24. 1. BC would be equala A 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 24. 1. would be less than the base EF; but it is. B D E not; therefore the angle BAC is not less than the angle EDF; and it was shown 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. PROP. XXVI. THEOR. Ir two triangles have two angles of 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. e; 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. AB to DE, and AC to DF, and the third angle BAC to the third G A D For, if AB be B E 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 sides GB, BC are equal to the two DE, BOOK I. EF, each to each; and the angle GBC is equal to the an gle DEF; therefore the base GC is equal to the base DF, * 4. 1. 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 DFÉ; 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 which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this case, the othersides 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 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 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 opposite angle BCA; which is impossible; wherefore BC is not unequal to EF, that is, it is 16. 1. 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. Boos I. PROP. XXVII. THEOR. Ir a straight line falling upon two other straight lines 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 ED. 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 * 16. 1. greater than the interior and opposite angle EFG; may be demonstrated, that C; but those straight lines which meet neither way, though 35 Def. produced ever so far, are parallel to one another. AB therefore is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. PROP. XXVIII. THEOR. Ir 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. с a 15. 1. Because the angle EGB is equal to the angle GHD, and Book I. the angle EGB equal to the angle AGH, the angle AGH is equal to the angle GHD; and they are the alternate angles; therefore AB is parallel to CD. Again, because ↳ 27. 1. the angles BGH, GHD are equal to two right angles; and By Hyp. that AGH, BGH are also equal to two right angles; the 13. 1. angles AGH, BGH, are equal to the two 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. PROP. XXIX. THEOR. notes on If a straight line fall upon two parallel straight straight See the lines, it makes the alternate angles equal to one this propo another; and the exterior angle equal to the inte- sition. rior 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, GHD, are equal to one another; and the exterior angle EGB is equal to the interior and opposite, upon the E same side GHD; and the two interior angles BGH, GHD upon the same side, are toge- A ther equal to two right angles. For, if AGH be not equal to GHD,one of them must be greater than the other; let AGH be the greater; and because the B D F a angle AGH is greater than the angle GHD, add to each of them the angle BGH; therefore the angles AGH, BGH are greater than the angles BGH, GHD; but the angles. AGH, BGH, are equala to two right angles; therefore the 13. 1. angles 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 side less than two right angles, do meet* together if continually 12 Ax. produced; therefore the straight lines AB, CD, if produced See the far enough, shall meet; but they never meet, since they are this propoparallel by the hypothesis; therefore the angle AGH is sition. not unequal to the angle GHD, that is, it is equal to it; * notes on b.15. 1. Book I. but the angle AGH is equal to the angle EGB; therefore likewise EGB is equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH are equal 13. 1. to the angles BGH, GHD; but EGB, BGH are equale to two right angles; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight, &c. Q.E. D. PROP. XXX. THEOR. STRAIGHT lines which are parallel to the same straight line are parallel to each other. Let AB, CD be each of them parallel to EF; AB is also parallel to CD. A Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the an* 29. 1. gle AGH is equal to the angle GHF. Again, because the straight line GK cuts the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD; and it was shown that the angle AGK is equal to the angle GHF; b E C -B H -F K -D therefore also AGK is equal to GKD; and they are alter27. 1. nate angles; therefore AB is parallelb to CD. Wherefore straight lines, &c. Q. E. D. PROP. XXXI. PROB. To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line; it is required to draw a straight line through the point A, paral- In BC take any point D, and E B D C 23. 1. the straight line AD, makea 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, |