*23.1. 3. 1. + Hyp. + Constr. 4.1. 5.1. † 9 Ax. Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF: the base BC shall be greater than the base EF. Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make the angle EDG equal to the angle BAC; and make DG equal to AC or DF, and join EG, GF. * Because AB is equal† to DE, and AC+ to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is D E G F + Constr. equal to the angle EDG; therefore the base BC is equal to the base EG. And because DG is equal to DF, the angle DFG is equal to the angle DGF; but the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than EGF; therefore much more is the angle EFG greater than the angle EGF and because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle is subtended by the greater side; therefore the side EG is greater than the side EF: but EG was proved to be equal to BC; therefore BC is greater than EF. Therefore, if two triangles, &c. Q. E. D. 19. 1. See N. * 4, 1. PROP. XXV. THEOR. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the base BC greater than the base EF: the angle BAC shall be greater than the angle EDF. For, if it be not greater, it must either be equal to it, or less than it: but the angle BAC is not equal to the angle EDF, because then the base BC would be equal * 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 would be less than the base EF; but it is+ not; therefore the angle BAC is not less than the angle EDF: and * D AA E it was shewn 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. 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 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. A G 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 sides shall be equal, each to each, viz. AB to DE, and AC to DF, and the third angle BAC to the third angle EDF. B CE F * + Hyp. * 24. 1. + Hyp. For, if AB be not equal to DE, one of them must be greater than the other. Let AB be the greater of the two, and make BG equal to DE, and join. GC; †3.1. therefore, because BG is equal to DE, and BC† to ↑ Hyp. EF, the two sides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is equal to + Hyp. the angle DEF; therefore the base GC is equal to the * 4. 1. base DF, 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 DFE: but DFE is, by the hypothesis, equal to the angle BCA; wherefore also the angle BCG is equal to the angle BCA, the +1 Ax. less to the greater, which is impossible: therefore AB + Hyp. + Hyp. * 4. 1. + 3. 1. + Hyp. + Hyp. + 4. 1. + Hyp. † 1 Ax. 16. 1. + Hyp. + Hyp. † 4. 1. 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; therefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. * HC E 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 other sides shall be equal, 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 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 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 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 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 containt 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. 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 it be not parallel, AB and CD being pro duced will meet either towards B, D, or towards A, C: A B G C: but those straight lines which meet neither way, though produced ever so far, are parallel * to one an other: 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 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 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: AB shall be parallel to CD. * -B -D F 16. 1. † Hyp. 35 Def + Hyp. * 15.1. †1 Ax. * 27.1. * Because the angle EGB is equal to the angle GHD, and the angle EGB is equal to the angle AGH, therefore the angle AGH is equal to the angle GHD: and they are the alternate angles; therefore AB is parallel * to CD. Again, because the angles BGH, GHD are equal to two right angles, and that AGH, BGH are By Hyp. also equal to two right angles; therefore the angles * 13. 1. AGH, BGH are equal to the angles BGH, GHD: †1 Ax. take away the common angle BGH; therefore the remaining angle AGH is equal † to the remaining angle † 3 Ax. GHD: and they are alternate angles; therefore AB is parallel to CD. Wherefore, if a straight line, &c. † 27.1. Q. E. D. * See the notes on this proposition. † 4 Ax. 13. 1. * 12 Ax. See the notes on PROP. XXIX. THEOR. 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 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, shall be equal to one another; and the exterior angle EGB shall be equal to the interior and opposite, upon the same side, GHD; and the two interior angles BGH, E GHD, upon the same side shall be together 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 B D F * because the 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 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 angles, will meet together if continually produced; therefore the straight lines AB, CD, if produced this propo- 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 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 † to the angles BGH, GHD: but EGB, BGH are equal to two right angles; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. sition. * 15. 1. † 1 Ax. † 2 Ax. 13. 1. † 1 Ax. * * 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 shall be parallel to CD. |