Let ABC, DEF, be two triangles, which have The two sides AB, AC equal to the two sides DE, DF, each to each-namely, AB 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. PROOF. For if the angle BAC be not greater than the A angle EDF, it must either be equal to it or less. But the angle BAC is not equal to the angle EDF, for then the base BC would be equal to the base EF (I. 4), but B it is not (Hyp.); Therefore the angle BAC is not equal to the angle EDF; Neither is the angle BAC less than the angle EDF, for then the base BC would be less than the base EF (I. 24), but it is not (Hyp.), < BAC not Therefore the angle BAC is not less than the angle EDF. EDF. And it has been proved that the angle BAC is not equal to the angle EDF; Therefore the angle BAC is greater than the angle EDF. Proposition 26.-—Theorem. 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-namely, either the side adjacent to the equal angles in each, or the side opposite to them; then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other. Or, If two angles and a side in one triangle be respectively equal to two angles and a corresponding side in another triangle, the triangles shall be equal in every respect. Let ABC, DEF be two triangles, which have The angles ABC, BCA equal to the angles DEF, EFD, each to each—namely, ABC to DEF, and BCA to EFD; Also one side equal to one side. Given CASE 1.-First, let the sides adjacent to the equal angles EF. in each be equal-namely, BC to EF; Suppose Make .. 4 GCB = 4 DFE. AB not unequal to DE. Suppose Make BH = EF. Then shall the side AB be equal to DE, the side AC to DF, and the angle BAC to the angle EDF. A B For if AB be not equal to DE, one of them must be D C E greater than the other. Let AB be the greater of the two. CONSTRUCTION.-Make BG equal to DE (I. 3), and join GC. PROOF.-Because BG is equal to DE (Const.), and BC F is equal to EF (Hyp.), the two sides GB, BC are equal to the two sides DE, EF, each to each. And the angle GBC is equal to the angle DEF (Hyp.); 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 (I. 4). But the angle DFE is equal to the angle BCA (Hyp.); Therefore the angle GCB is equal to the angle BCA (Ax. 1), 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 (Hyp.); Therefore the two sides AB, BC are equal to the two sides And the angle ABC is equal to the angle DEF (Hyp.); A B Likewise in this case the other sides shall be equal, AC to H C B DF, and BC to EF; and also the angle BAC to the angle EDF. For if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater of the two. equal to EF (I. 3), and join AH, CONSTRUCTION.Make BH PROOF.-Because BH is equal to EF (Const.), and AB is BH EF, equal to DE (Hyp.), the two sides AB, BH are equal to the AB DE two sides DE, EF, each to each, And the angle ABH is equal to the angle DEF (Hyp.); 4 ABH = Z DEF. .. < BHA = EFD Therefore the angle BHA is equal to the angle EFD (I. 4). But the angle EFD is equal to the angle BČA (Hyp.); Therefore the angle BHA is also equal to the angle BCA BCA. (Ax. 1); That is, the exterior angle BHA of the triangle AHC, is equal to its interior and opposite angle BCA, which is impossible (I. 16); Therefore BC is not unequal to EF—that is, it is equal to it; BC not and AB is equal to DE (Hyp.); Therefore the two sides AB, BC are equal to the two sides And the angle ABC is equal to the angle DEF (Hyp.); And the third angle BAC is equal to the third angle Therefore, if two triangles, &c. Q. E. D. Proposition 27.-Theorem. 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_A AEF, EFD, equal to one another. AB shall be parallel to CD. For if AB and CD be not parallel, they will meet if produced, either towards B, D, or towards A, C. C F E unequal to EF. B Given G < EFD. Let them be produced, and meet towards B, D, in the point G. < EFG, <AEF> PROOF. Then GEF is a triangle, and its exterior angle AEF is greater than the interior and opposite angle EFG (I. 16). and also = 4 EFG. 2 AGH= < GHD. But the angle AEF is also equal to EFG (Hyp.), which is impossible; Therefore AB and CD, being produced, do not meet towards B, D. In like manner it may be shown that they do not meet towards A, C. But those straight lines in the same plane which being produced ever so far both ways do not meet are parallel (Def. 34); Therefore AB is parallel to CB. Therefore, if a straight line, &c. Q. E. D Proposition 28.-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, 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; E A Or make the interior angles on the -B same side, the angles BGH, GHD, together equal to two right angles; AB shall be parallel to CD. PROOF 1.-Because the angle EGB F is equal to the angle GHD (Hyp.), And the angle EGB is equal to the angle AGH (I. 15); Therefore the angle AGH is equal to the angle GHD (Ax. 1), and these angles are alternate; Therefore AB is parallel to CD (I. 27). PROOF 2.-Again, because the angles BGH, GHD are equal to two right angles (Hyp.), And the angles BGH, AGH are also equal to two right angles (I. 13). Therefore the angles BGH, AGH are equal to the angles BGH BGH, GHD (Ax. 1). Take away the common angle BGH. Therefore the remaining angle AGH is equal to the remain- Proposition 29.-Theorem. 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 also 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. The exterior angle EGB shall be equal to GHD, the interior and opposite angle upon the same side; And the two interior angles on the same side BGH, GHD shall be together equal to two right angles. For if AGH be not equal to GHD, one of them must be greater A E than the other. Let AGH be the greater. -B PROOF.-Then the angle AGH is greater than the angle GHD; to each of them add the angle BGH. Therefore the angles BGH, AGH are greater than the angles BGH, GHD (Ax. 4). But the angles BGH, AGH are together equal to two right anglès (1. 3). + AGH = BGH +4 GHD. = 4 GHD. < AGH > GIID. (suppose.) .. 4 BGH Therefore the angles BGH, GHD are less than two right+ GHD angles. But if a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these straight lines being <two right angles. |