CONSTRUCTION For if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater, make BH equal to EF, (1. 3) and join AH. DEMONSTRATION Then, in the two triangles ABH, DEF, because BH is assumed to be equal to EF, and AB is equal to DE, (hyp.) the two sides AB, BH, must be equal to the two DE, EF, each to each; and the angle ABH is equal to the angle DEF; (hyp.) therefore the base AH must be equal to the base DF, and the triangle ABH to the triangle DEF, and the remaining angles of the one equal to the remaining angles of the other, each to each, to which the equal sides are opposite; therefore the angle BHA must be equal to the angle BCA, (1. 4) that is, the exterior angle, BHA, of the triangle ABC, must be equal to its interior and opposite angle BCA, which is impossible; (1. 16) wherefore BC is not unequal to EF, that is, BC is equal to EF. Therefore in the triangles ABC, DEF, because AB is equal to DE, (hyp.) and BC has been shown to be equal to EF, the two AB, BC, are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; (hyp.) therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. (1. 4.) Wherefore, if two triangles, &c. Q. E. D. PROP. XXVII.- THEOREM. If a straight line, falling upon two other straight lines, make the alternate angles equal to one another, then these two straight lines shall be parallel. (References-Prop. 1. 16; def. 35.) Let the straight line EF, which falls upon the two straight lines, AB, CD, make the alternate angles, AEF, EFD, equal to one another. Then AB shall be parallel to CD. 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. and its exterior angle AEF must be greater than the interior and opposite angle EFG; (1. 16) but the angle AEF is equal to the angle EFG; (hyp.) therefore the angle AEF must be both greater than, and equal to the angle EFG ; which is impossible. Therefore AB, CD, being produced, do not meet towards B, D. In like manner it may be demonstrated that AB, CD, do not meet towards A, C. But those straight lines, in the same plane, which meet neither way, though produced ever so far, are parallel to one another; (def. 35) therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. PROP. XXVIII.-THEOREM. 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; then the two straight lines shall be parallel to one another. (References Prop. 1. 13, 15, 27; ax. I. 3.) Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD upon the same side; or make the interior angles on the same side BGH, GHD, together equal to two right angles. Then AB shall be parallel to CD. E 'B DEMONSTRATION Because the angle EGB is equal to the angle GHD, (hyp.) and the angle EGB is equal to the angle AGH, (1. 15) therefore the angle AGH is equal to the angle GHD; (ax. 1) and they are alternate angles, therefore AB is parallel to CD. (1. 27) Again, because the angles BGH, GHD, are equal to two right angles; (hyp.) and that the angles AGH, BGH, are also equal to two right angles; (1. 13) therefore the angles AGH, BGH, are equal to the angles BGH, GHD; (ax. 1) take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; (ax. 3) and they are alternate angles; therefore AB is parallel to CD. (1. 27) Wherefore, if a straight line, &c. Q. E. D. PROP. XXIX.- THEOREM. If a straight line fall upon two parallel straight lines, then it makes the alternate angles equal to each other; and the exterior angle equal to the interior and opposite angle upon the same side; and likewise the two interior angles upon the same side together equal to two right angles. (References-Prop. I. 13, 15; ax. I. 2, 4, 12.) Let the right line EF, fall on the parallel lines AB, CD. Then the alternate angle AGH shall be equal to the alternate angle GHD; and the exterior angle EGB shall be equal to the interior and opposite angle GHD, on the same side of the line EF; and the two interior angles BGH, GHD, on the same side, shall be together equal to two right angles. Then because the angle AGH is assumed to be greater than the angle GHD, add to each of them the angle BGH; therefore the angles AGH, BGH, must be greater than the angles BGH, GHD; (ax. 4) but the angles AGH, BGH, are equal to two right angles; (1. 13) therefore the angles BGH, GHD, must be 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, will meet together if continually produced;' (ax. 12) therefore the straight lines AB, CD, must meet if produced far enough. but they never meet, since they are parallel by the hypothesis; therefore the angle AGH is not unequal to the angle GHD, that is, the angle AGH is equal to the angle GHD, which are the alternate angles. But the angle AGH is equal to the angle EGB; (1. 15) therefore the angle EGB is equal to the angle GHD, (ax. 1) that is, the exterior angle is equal to the interior and opposite. Add to each of these the angles BGH; therefore the angles EGB, BGH, are equal to the angles BGH, GHD; (ax. 2) but EGB, BGH are equal to two right angles; (1. 13) therefore also BGH, GĦD, are equal to two right angles, (ax. 1) which are the two interior angles. Wherefore, if a straight line, &c. PROP. XXX.—THEOREM. Q. E. D. Straight lines which are parallel to the same straight line are parallel to one another. (References-Prop. I. 27, 29; ax. 1.) Let AB, CD, be each of them parallel to EF. Then AB shall be also parallel to CD. Because GHK cuts the parallel straight lines AB, EF, therefore the angle AGH is equal to the angle GHF. (1. 29.) Again, because the straight line GHK cuts the parallel straight lines EF, CD, therefore the angle GHF is equal to the angle GKD; and it was shown that the angle AGK is equal to the angle GHF; therefore also the angle AGK is equal to the angle GKD; (ax. 1) and they are alternate angles; therefore AB is parallel to CD. (1. 27.) Wherefore, straight lines, &c. Q. E. D. |