* 1 Ax. that are equal to the same thing, are equal * to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC: but CBE, EBD are two right angles; therefore, DBA, ABC are together equal † to two right angles. Wherefore, when a straight line, &c. +1 Ax. Q. E. D. PROP. XIV. THEOR. E If, at a point, in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles At the point B in the straight line For, if BD be not in the same straight line with CB, let BE be in the same straight line with it: therefore, because the straight line AB makes with the straight line CBE, upon one side of it, the angles ABC, ABE, these angles are together equal* to two right angles; but the angles ABC, ABD are likewise together equal + to two right angles; therefore, the angles CBA, ABE are equal † to the angles CBA, ABD: take away the common angle ABČ, and the remaining angle ABE is equal * to the remaining angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with BC. And, in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which, therefore, is in the same straight line with CB. Wherefore, if at a point, &c. Q. E. D. * 13. 1. + Hyp. † 1 Ax. * 3 Ax. PROP. XV. THEOR. If two straight lines cut one another, the vertical, or op posite, angles shall be equal. Let the two straight lines AB, CD, cut one another in the point E: the angle AEC shall be equal to the angle DEB, and CEB to AED. Because the straight line AE makes with CD the angles CEA, AED, these angles are together equal * to * 13. 1. * 13. 1. A D two right angles. Again, because the straight line DE makes with AB the angles AED, DEB, these also are together equal to two right angles; and CEA, AED have been demonstrated to be equal to two right angles; wherefore the angles CEA, AED are equal † to the angles AED, DEB, 11 Ax. Take away the common angle AED, and the remaining angle CEA is equal * to the remaining angle DEB. *3 Ax. In the same manner it can be demonstrated, that the angles 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 where they cut, are together equal to four right angles. Cor. Ž. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles. PROP. XVI. THEOR. * 10, 1. + 3. 1. + Constr. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let its side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC. Bisect* AC in E, join BE and produce it to F, and make EF equalt to BE, and join FC. Because AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; and the angle B AEB, is equal* to the angle CEF, because they are opposite vertical angles; therefore the base AB is equal* to the base CF, and the triangle AEB to the *4. 1. 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 + 9 AX. angle ECF, therefore the angle ACD is greater than BĂE. In the same manner, if the side BC be bisected, and AC be produced to G, it may be demonstrated * 15. 1. с * 15. 1. that the angle BCG, that is, the angle* ACD, is greater than the angle ABC. Therefore, if one side, &c. Q. E. D. PROP. XVII. THEOR. A * 16, 1. Any two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles together shall be less than two right angles. Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater * than the interior and opposite angle ABC; to each of these add the angle ACB; therefore D the angles ACD, ACB are greatert than the angles ABC, ACB: but ACD, ACB are together equal* 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. B с + 4 Ax. * 13. 1. Q. E. D. PROP. XVIII. THEOR. А D 3.1. The greater side of every triangle is opposite to the greater angle. Because AC is greater than AB, * 16.1. * 5.1. + Constr. 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 shall be greater than the side ÅB. 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* to the angle CAB; but it * 5. 1. ist not; therefore AC is not equal to + Hyp. AB: neither is it less; because then the angle ABC would be less 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. Wherefore the greater angle, &c. Q. E. D. * 18. 1. B PROP. XX. THEOR. D * 3.1. B * 5. 1. Any two sides of a triangle are together greater than the See N. third side. Let ABC be a triangle: any two sides of it together shall be 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* AD equal to AC; and join DC. Because DA is equal to AC, the angle ADC is equal* to ACD; but the angle BCD is greatert than the 19 Ax. angle ACD; therefore the angle BCD is greater than the angle ADC: and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater* angle is subtended by the greater * 19.1. side; therefore the side DB is greater than the side sides BA, AC araqual to BA and AC; therefore the 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. PROP. XXI. THEOR. two straight lines to a point within the triangle, these 20. 1. B + 4 Ax. + 20. 1. the ends of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle: BD and DC shall be less than the other two sides BA, AC of the triangle, but shall contain an angle BDC greater than the angle BAC. Produce BD to E; and because two sides of a triangle* 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, add DB to each of these; therefore the sides CE, EB, are greatert 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. Again, because the exterior angle of a triangle * 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 CEB; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q. E. D. + 4 Ax. * 16. 1. PROP. XXII. PROB. See N. * 20. 1. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third*. 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 greater 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 termi- K HE * 3.1. * 3 Post. |