Then the angle BAC shall be greater than the angle EDF. Demonstration. For, if the angle BAC be not greater than the angle EDF, it must either be equal to it or less than it. If the angle BAC were equal to the angle EDF, then the base BC would be equal to the base EF (I. 4); but it is not equal (hyp.), therefore 1. The angle BAC is not equal to the angle EDF. Again, if the angle BAC were less than the angle EDF, then the base BC would be less than the base EF (I. 24), but it is not less (hyp.), therefore 2. The angle BAC is not less than the angle EDF, and it has been shown, that the angle BAC is not equal to the angle EDF; therefore 3. The angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q.E.D. 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-viz., either the sides adjacent to the equal angles in each, or the sides 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. Ал 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. B E F First, let those sides be equal which are adjacent to the angles that are equal in the two triangles, namely, BC to EF. Then the other sides shall be equal, each to each, namely, AB to DE, and AC to DF, and the third angle BAC to the third angle EDF. Construction. For, if AB be not equal to DE, one of them must be greater than the other. If possible, let AB be greater than DE, make BG equal to ED (I. 3), and join GC. Demonstration. Then in the two triangles GBC, DEF, because GB is equal to DE, and BC to EF (hyp.), 1. 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 2. The base GC is equal to the base DF, and the triangle GBC to the triangle DEF (I. 4), and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore 3. The angle GCB is equal to the angle DFE; but the angle ACB is, by the hypothesis, equal to the angle DFE; wherefore also 4. The angle GCB is equal to the angle ACB (Ax. 1) ; the less angle equal to the greater, which is impossible; therefore 5. AB is not unequal to DE, that is, AB is equal to DE. Hence in the triangles ABC, DEF; because AB is equal to DE, and BC to EF (hyp.), and the angle ABC is equal to the angle DEF (hyp.), therefore 6. The base AC is equal to the base DF (I. 4), and the third angle BAC to the third angle EDF. Secondly, let the sides which are opposite to one of the equal angles in each triangle be equal to one another, namely, AB equal to DE. Then in this case likewise, the other sides shall be equal, AC to DF, and BC to EF, and also the third angle BAC to the third angle EDF. B H E F Construction. For if BC be not equal to EF, one of them must be greater than the other. If possible, let BC be greater than EF, make BH equal to EF (I. 3), and join AH. Demonstration. Then in the two triangles, ABH, DEF, because AB is equal to DE, and BH to EF, 1. The two sides AB, BH are equal to the two DE, EF, to each, and the angle ABH is equal to the angle DEF (hyp.), therefore each 2. The base AH is equal to the base DF, and the triangle ABH to the triangle DEF (I. 4), and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore 3. The angle BHA is equal to the angle EFD, but the angle EFD is equal to the angle BCA (hyp.), therefore 4. The angle BHA is equal to the angle 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); wherefore 5. BC is not unequal to EF, that is, BC is equal to EF. Hence in the triangles ABC, DEF because AB is equal to DE, and BC to EF (hyp.), and the included angle ABC is equal to the included angle DEF (hyp.), therefore 6. The base AC is equal to the base DF (I. 4), and the third angle BAC to the third angle EDF. Wherefore, if two triangles, &c. Q.E.D. PROPOSITION 27.-Theorem. If a straight line falling on two other straight lines, make the alternate angles equal to each other; 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 ÂEF, EFD, equal to one another. Then AB shall be parallel to CD. E D Construction. For, if AB be not parallel to CD, then AB and CD being produced will meet, either towards A and C, or towards B and D. Let AB, CD be produced and meet, if possible, towards B and D in the point G. Demonstration. Then 1. GEF is a triangle, and its exterior angle A EF is greater than the interior and opposite angle EFG (I. 16). but the angle AEF is equal to the angle EFG (hyp.), therefore the angle AEF is greater than and equal to the angle EFG, which is impossible; therefore 2. AB and CD being produced do not meet towards B, In like manner it may be demonstrated, that D. 3. AB and CD being produced 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 4. AB is parallel to CD. Wherefore, if a straight line, &c. QE.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 of the line EF; or make the two interior angles BGH, GHD on the same side, together equal to two right angles. Then AB shall be parallel to CD. E A H D Demonstration. Because the angle EGB is equal to the angle GHD (hyp.) and the angle EGB is equal to the angle AGH (I. 15), therefore 1. The angle AGH is equal to the angle GHD (Ax. 1), and they are alternate angles; therefore 2. AB is parallel to CD (I. 27). Again, because the angles BGH, GHD are together equal to two right angles (hyp.), and that the angles AGH, BGH are also together equal to two right angles (I. 13), therefore 1. The angles AGH, BGH are equal to the angles BGH, GHD (Ax. 1); take away from these equals the common angle BGH, therefore 2. The remaining angle AGH is equal to the remaining angle GHD (Ax. 3), and they are alternate angles; therefore 3. AB is parallel to CD (I. 27). Wherefore, if a straight line, &c. Q.E.D. 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 likewise the two interior angles upon the same side together equal to two right angles. |