* 24. 1. but it is not; therefore the angle BAC is not equal to the angle EDF; neither is it less; because then the base BC would be less than the base EF; but it is not; therefore the angle BAC is not less than the angle EDF; and it was proved that it is not equal to it; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. PROP. XXVI. THEOR. Q. E. D. If two triangles have two angles of the 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 the equal angles in each; then shall the other sides be equal, each to each; and the third angle of the one equal to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, each to each, 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 G B D CE F For, if AB be not equal to DE, one of them must be greater than the other. Let AB be the greater, and make BG equal to ED, and join GC; therefore, because BG is equal to ED, and BC to EF, the two sides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is equal to the angle DEF; therefore the triangle GBC is equal 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* DFE; but DFE is, * 1 Ax. by the hypothesis, equal to the angle ACB; 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 sides AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; therefore the base AC is equal to the base DF, and the third angle BAC * 4. 1. to the third angle EDF. Next, let the sides which are opposite to the equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this case, the other sides shall be equal, B BCE viz. AC to DF, and BC to EF; and also the third angle BAC to the third angle EDF. For, if BC be not equal to EF, let BC be the greater, and make BH equal to EF, and join AH; then, be- * 3. 1. cause BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two DE, EF, each to each; and they contain equal angles; therefore the triangle ABH is equal to the triangle DEF, and the other * 4. 1. angles to the other angles, 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 the angle BHA is also equal* to * 1 Ax. the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle HCA; which is impossible:* wherefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE; therefore the two sides 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 *4. 1. D 2 16. 1. DF, and the third angle BAC to the third angle EDF. PROP. XXVII. THEOR. 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 the straight line EF, which falls upon the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another; AB shall be parallel to CD. For, if AB be not parallel to CD, AB and CD being produced will 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 ex16. 1. terior angle AEF is greater than the interior and opposite angle EFG; but this is impossible, because the wards B, D. In like manner it may be demonstrated that they do not meet towards A, C; but those straight lines which do not meet, though produced either way * 35 Def. continually, are parallel to one another: therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. PROP. XXVIII. THEOR. If 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 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 straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB A G equal to the interior and opposite angle GHD upon Because the angle EGB is equal* * Hyp. to the * 15. 1. Hyp. 13. 1. to the angle GHD, and the angle EGB equal angle AGH, therefore the angle AGH is equal to the angle GHD; and they are alternate angles; therefore AB is parallel* to CD. Again, because the angles * 27. 1. BGH, GHD are together equal to two right angles, and AGH, BGH, are also together equal to two right angles; therefore the angles AGH, BGH are equal* to the angles BGH, GHD: take away the common angle BGH; and the remaining angle AGH is therefore equal to the remaining angle GHD; and they are 3 Ax. alternate angles: therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. If a straight line fall upon two parallel straight lines, it B side; and the two interior an- A. gles BGH, GHD upon the same side, shall be together equal to two right angles. For, if AGH be not equal to C G A * 1 Ax. * 13. 1. GHD, one of them must be greater than the other: let AGH be the greater; and because the angle AGH is greater than the angle GHD, add to each of these unequals the angle BGH; therefore the angles AGH, * 4 Ax. BGH are greater* than the angles BGH, GHD; but the angles AGH, BGH are together equal to two right angles; therefore the 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 * 12 Ax. angles meet,* if far enough produced; therefore the straight lines AB, CD, if produced far enough, will meet; but they never meet, since they are parallel by the hypothesis; therefore the angle AGH is not unequal to the angle GHD, that is, it is equal to it: but *15. 1. the angle AGH is equal* to the angle EGB; therefore EGB is also equal to GHD: add to each of these equals the angle BGH; then will the angles EGB, BGH be equal to the angles BGH, GHD; but EGB, BGH are equal to two right angles; therefore BGH, GHD are also equal to two right angles. Wherefore, if a straight, &c. Q. E. D. * 2 Ax. * 13. 1. * 1 Ax. 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 shall be parallel to CD. Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle * 29. 1. AGH is equal to the alternate angle GHF. Again, because the straight line GK cuts the parallel straight lines EF, CD, the angle 29. 1. GHF is equal to the angle GKD; * A E H C K G B |