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 = int. / GHD, Prop. 12. but AGH+ / BGH= 2 rt. ≤ s, s. Ax. 12. .. Hyp. BGH+≤ GHD<2 rt. AB, CD, will meet if prod. far enough; but they never meet, since they are parallel; AGH=/ GHD, LAGH is not Prop. 14. i. e. but < AGH=/ EGB, Ax. 1. GHD, Ax. 2. EGB=GHD; add to each the BGH, ../EGB+ BGH=/ BGH+GHD, Prop. 12. but .. EGB+ / BGH = 2 rt. s, BGH+ ≤ GHD= 2 rt. s. Wherefore, if a straight line, &c. 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 Let str. line GK cut AB, EF, CD, in the points G, H, K. Then GK cuts the lines AB, EF, Prop. 28. Again, ... GK cuts the || lines EF, CD, Prop. 28. F; Ax. 1. and it was proved / AGH = / GHF / AGH = / GKD, which are alt. s. ... 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, Z ADC, then shall EF || BC. Therefore, a str. line has been drawn through A || BC. Prop. 26. Prop. 22. Prop. 26. PROP. XXXI. THEOR. 32. 1 Eu. If a side of any triangle be produced, the ex- Let ABC be a A, and the side BC be BCA + ≤ CAB = 2 rt. ▲ s. Ax. 2 Ax. 2. Draw CE || AB; AC meets them, alt. BAC alt. / ACE. :: BD falls upon the lines AB, CE, ={ Z ABC, the int. and oppo. but LACE = BAC. ..whole ext. ACD = {BAC + < to each of these equals add ACB, LACB+2 ABC; but ▲ ACD+≤ ACB = 2 rt. ▲ s. :: ≤ ACB + ▲ BAC+ / ABC = 2 rt. ▲ s. Wherefore, if a side of a triangle, &c. COR. 1.-All the int. figure, together with 4 rt. twice as many rt. s as the Prop. 12. Ax. 1. s of any rectilin. C B For any rectilin. figure ABCDE can be divided into as many As as the figure has sides, by drawing str. lines from a pt. F within the figure to each of its angles. Then, by the preceding proposition, all the s of theses are equal to twice as many rt. s as there are As, i. e. as there are sides of the figure: and the same s are equal to the s of the figure, together with the s at the pt. F, which is the common vertex of the As; that is, together with 4 rt. s. COR. 2.-All the ext. s of any rectilin. figure are together equal to 4 rt. s. Cor. 2. |