PROP. XXVI. THEOR. 27. 1 Eu. 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 str. line EF, which falls upon str. lines AB, CD, make alt. EFD: then shall AB || CD. AEF the alt. For if ABCD, they will, when produced, meet either towards B, D, or towards A, C. Suppose the former, and let them meet in G, then GEF will form a but ; ..ext. AEF > int. ▲ EFG; AEF= Def. 35. Prop. 15. EFG or EFD, Hyp. which is impossible. ... AB, CD, being produced, do not meet to wards B, D. In like manner it may be proved that they Wherefore if a str. line, &c. Def. 35. Ꭰ PROP. XXVII. THEOR. 28. 1 Eu. If a straight line falling upon two other straight lines, make the exterior angle equal to the interior and opposite upon the same side of the line; or make the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another. Let the str. line EF, which falls upon AB and CD, make ext. / EGB=GHD, the int. and oppo. on the same side. Or int.s on the same side, viz. BGH+ GHD 2 rt. Ls; PROP. XXVIII. THEOR. 29. 1 Eu. 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 angle on the same side; and likewise the two interior angles upon the same side together equal to two right angles. Let the str. line EF fall on the || str. lines AB, CD, then alt. ▲ AGH = alt. / GHD, ext. EGB = And BGH+ / E int. / GHD, GHD = 2 rt. s. Prop. 12. but AGH+ / BGH= 2 rt. ≤ s, Ax. 12. Hyp. BGH+ ≤ GHD<2 rt. AB, CD, will meet if prod. far enough; but they never meet, since they are parallel; LAGH is not GHD, s. / AGH=/ EGB, EGB=GHD; Ax. 2. :: Prop. 12. but .. <EGB+/ BGH=/ BGH+▲ GHD, EGB+ / BGH= 2 rt. ≤ s, BGH+ ≤ GHD= 2 rt. Wherefore, if a straight line, &c. s. Prop. 28. PROP. XXIX. THEOR. 30. 1 Eu. Straight lines which are parallel to the same straight line, are parallel to each other. Let AB, CD, be each || EF; then shall AB || CD. Let str. line GK cut AB, EF, CD, in the points G, H, K. Then ... GK cuts the || lines AB, EF, Again, GK cuts the lines EF, CD, and it was proved ▲ AGH = ▲ GHF; which are alt. s. Prop. 28. Ax. 1. ... AB || CD. Wherefore str. lines, &c. PROP. XXX. PROB. 31. 1 Eu. To draw a straight line through a given point, parallel to a given straight line. Let A be the given point, BC the given str. line. To draw a str. line through A || BC. In BC take any pt. D, join AD, make DAE = prod. EA to F, ▲ ADC, then shall EF || BC. Therefore, a str. line has been drawn through A || BC. Prop. 26. Prop. 22. Prop. 26. |