angle DCB; but ADB is equal (5. 1.) to ABD, because the side AB is equal to the side AD; therefore the angle ABD is likewise greater than the angle ACB; wherefore much more is the angle ABC greater than ACB PROP. XIX. THEOR. The greater angle of every triangle is subtended by the greater side, or hus the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the side AC is likewise greater than the side AB. For, if it be not greater, AC must either be equal to AB, or less than it; it is not equal, because then the angle ABC would be equal (5. 1.) to the angle ACB; but it is not; therefore AC is not equal to AB; neither is it less; because then the angle ABC would be less (18. 1.) than the angle ACB; but it is not; therefore the side AC is not less than AB; and it has been shewn that it is not equal to AB; therefore AC is greater than AB. A B PROP. XX. THEOR. Any two sides of a triangle are together greater than the third side. Let ABC be a triangle; any two sides of it together are greater than the third side, viz. the sides BA, AC greater than the side BC; and AB, BC greater than AC; and BC, CA greater than AB. gle BDC, and that the greater (19. 1.) side is opposite to the greater angle; therefore the side DB is greater than the side BC; but DB is equal to BA and AC together; therefore BA and AC together are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB. SCHOLIUM. This may be demonstrated without producing any of the sides: thus, the line BC, for example, is the shortest distance from B to C; therefore BC is less than BA+AC or BA+AC>BC. PROP. XXI. THEOR. If from the ends of one side of a triangle, there be drawn two straight lines to a point within the triangle, these two lines shall be less than the other two sides of the triangle, but shall contain a greater angle. Let the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC. A D E Produce BD to E; and because two sides of a triangle (20. 1.) are greater than the third side, the two sides BA, AE of the triangle ABE are greater than BE. To each of these add EC; therefore the sides BA, AC are greater than BE, EC: Again, because the two sides CE, ED, of the triangle CED are greater than CD, if DB be added to each, the sides CE, EB, will be greater than CD, DB; but it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC. B C Again, because the exterior angle of a triangle (16. 1.) is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC is greater than the angle CEE; much more then is the angle BDC greater than the angle BAC. PROP. XXII. PROВ. To construct a triangle of which the sides shall be equal to three given straight lines; but any two whatever of these lines must be greater than the third (20. 1.). 1 the centre F, at the distance FD, describe (3. Post.) the circle DKL, and from the centre G, at the distance GH, describe (3. Post.) another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines, A, B, C. Because the point F is the centre of the circle DKL, FD is equal (11. Def.) to FK; but FD is equal to the straight line A; therefore FK is equal to A: Again, because G is the centre of the circle LKH, GH is equal (11. Def.) to GK; but GH is equal to C; therefore, also, GK is equal to C; and FG is equal to B; therefore the three straight lines KF, FG, GK, are equal to the three A, B, C: And therefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines, A, B C. SCHOLIUM. If one of the sides were greater than the sum of the other two, the arcs would not intersect each other: but the solution will always be possible, when the sum of two sides, any how taken (20. 1.) is greater than the third. PROP. XXIII. PROв. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle; it is required to make an angle at the given point A in the given straight line AB, that shall be equal to the given rectilineal angle DCE. Take in CD, CE any points D, E, and join DE; and make (22. 1.) the triangle AF, the sides of which shall be equal to the three straight lines, CD, DE, СЕ, so that CD be equal to AF, CE to AG, and DE to FG; and because DC, CE are equal to FA, AG, D each to each, and the base DE to the base FG; the angle DCE is B equal (8. 1.) to the angle FAG. Therefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. PROP. XXIV. THEOR, If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides of the other; the base of that which has the greater angle shall be greater than the base of the other. 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 1 DF; but the angle BAC greater than the angle EDF; the base BC is also 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 (23. 1.) the angle EDG equal to the angle BAC: and make DG equal (3. 1.) 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 equal to the angle EDG, therefore the base BC is equal (4.1.) to the base EG; and because DG is equal to DF, the angle DFG is equal (5. 1.) to the angle DGF; but the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than EGF; and 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 because the greater (19.1.) side is opposite to the greater angle, the side EG is greater than the side EF; but EG is equal to BC; and therefore also BC is greater than EF. 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 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 let the base CB be greater than the base EF, the angle BAC is likewise greater than the angle EDF. For, if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle EDF, because then the base BC would be equal (4.1.) 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 (24. 1.) than the bas EF; but it is not; therefore the angle BAC is not less than the angle EDF: and it was shewn that it is not equal to it: therefore the angle BAC is greater than the angle EDF. В 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 sidesbe equal, each to each; and also the third angle of the one 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, viz. ABC to DEF, and BCA to EFD, also one side equal to one side; and first, let those sides be equal which are A G D adjacent to the angles that are B For, if AB be not equal to DE, one of them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join GC; therefore, because BG is equal to DE, 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 base GC is equal (4. 1.) to the 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 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 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 (4. 1.) to the base DF, and the angle BAC to the angle EDF. 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 (4. 1.) |