For the angles CEA, AED, which the straight line AE makes with the straight line CD, are together equal (13. 1.) to two right angles and the angles AED, DEB, which the straight line DE makes with the straight line AB, are also together equal (13. 1.) to two right angles; A therefore the two angles CEA, AED are equal to the two AED, DEB. Take away * the common angle AED, and the remaining angle CEA is C B E D equal (3. Ax) to the remaining angle DEB. In the same manner it may be demonstrated that the angle CEB, AED are equal. Therefore, if two straight lines, &c. Q. E. D. COR. 1. From this it is manifest, that if two straight lines cut one another, the angles which they make at the point of their intersection, are together equal to four right angles. Cor. 2. And hence, all the angies made by any number of straight lines meeting in one point, are together equal to four right angles. PROP. XVI. THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior, and opposite angles. Let ABC be a triangle, and let its side BC be producea to D, the exterior angle ACD is greater than either of the interior opposite angles CBA, BAС. Bisect (10. 1.) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G. A F Because AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; and the angle AEB is equal (15. 1.) to the angle CEF, because they are vertical angles; B therefore the base AB is equal (4. 1.) to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ECD, that is ACD, is greater than BAE: In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, that is (15. 1.) the angle ACD, is greater than the angle ABC. Therefore if one side, &c. Q. E. D. PROP. XVII. THEOR. Any two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles together are less than two right angles. Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater (16. 1.) than the interior and opposite angle ABC; to each of these add the angle ACB; therefore the angles ACD, ACB are greater than the angles ABC, В ACB; but ACD, ACB are to gather equal (13. 1.) to two right angles; therefore the angles ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as also, CAB, ABC, are less than two right angles. Therefore, any two angles, &c. Q. E. D. PROP. XVIII. THEOR. The greater side of every triangle has the greater angle opposite to it. Let ABC be a triangle of which the side AC is greater than the side AB; the angle ABC is also greater than the angle BCA. From AC, which is greater than AB, cut off (3. 1.) AD equal to AB and join BD; and because ADB is the exterior angle of the triangle BDC, B it is greater (16. 1.) than the inte A D rior and opposite 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. Therefore the greater side, &c. Q.E. D. PROP. XIX. THEOR. The greater angle of every triangle is subtended by the greater side, or has 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. A 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 B 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. Wherefore the greater angle, &c. Q. E. D C 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. Produce BA to the point D, and make (3. 1.) AD equal to AC; and join DC. A C D Because DA is equal to AC, the angle ADC is likewise equal (5. 1.) to ACD; but the angle BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC; and B because the angle BCD of the triangle DCB is greater than its angle 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. Therefore any two sides, &c. Q. E. D. E 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. Produce BD to E; and because two sides of a triangle (20. 1.) are greater than the third side, the two greater than BD, DC. A E D 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 CЕВ; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q. E. D. 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.) Let A, B, C be the three given straight lines, of which any two whatever are greater, than the third, viz. A and B greater than C; A and C greater than B; and B and C than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each. Take a straight line DE, terminated at the point D, but unlimited towards E, and make (3. 1.) DF equal to A, FG to B, and GH equal to C; and from the centre F, D 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 E C 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. Which was to be done. 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 an gle at the given point A in the given straight line AB, that shall be equal to the given rec tilineal angle DCE. the three straight lines, CD, DE, CE so that CD be equal to Take in CD, CE any points D, E, and join DE; and make (22. 1.) the triangle AFG, the sides of which shall be equal to AF, CE to AG, and DE to FG; and because DC, CE are equal to FA, AG, each to each, and B the base DE to the base FG; the angle DCE is 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. Which was to be done. : |