Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. But the angles CBE, EBD have been proved equal to the same three angles; and things which are equal to the same thing are equal to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC; but the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are together equal to two right angles. (ax. 1.) Wherefore when a straight line, &c. PROPOSITION XIV. THEOREM. Q.E.D. If at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. At the point B in the straight line AB, let the two straight lines BC, BD upon the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles. Then BD shall be in the same straight line with CB. For, if BD be not in the same straight line with CB, let BE be in the same straight line with it. Then because AB meets the straight line CBE; therefore the adjacent angles CBA, ABE are equal to two right angles; (1. 13.) but the angles CBA, ABD are equal to two right angles; (hyp.) therefore the angles CBA, ABE are equal to the angles CBA, ABD: (ax.2.) take away from these equals the common angle CBA, therefore the remaining angle ABE is equal to the remaining angle ABD; (ax. 3.) the less equal to the greater angle, which is impossible: And in the same 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. PROPOSITION XV. THEOREM. If two straight lines cut one another, the vertical, or opposite angles shall be equal. Let the two straight lines AB, CD cut one another in the point E. Then the angle AEC shall be equal to the angle DEB, and the angle CEB to the angle AED. Because the straight line AE makes with CD at the point E, the adjacent angles CEA, AED; these angles are together equal to two right angles. (1. 13.) Again, because the straight line DE makes with AB at the point E the adjacent angles AED, DEB; these angles also are equal to two right angles; but the angles CEA, AED have been shewn to be equal to two right angles; wherefore the angles CEA, AED are equal to the angles AED, DEB; take away from each the common angle AED, and the remaining angle CEA is equal to the remaining angle DEB. (ax. 3.) In the same manner it may be demonstrated, that the angle CEB is equal to the angle AED. Q. E. D. Therefore, if two straight lines cut one another, &c. COR. 1. From this it is manifest, that, if two straight lines cut each other, the angles which they make at the point where they cut, are together equal to four right angles. COR. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles. 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. Then the exterior angle ACD shall be greater than either of the interior opposite angles CBA or BAC. Bisect AC in E, (1. 10.) and join BE; produce BE to F, making EF equal to BE, (1. 3.) and join FC. Because AE is equal to EC, and BE to EF; the two sides AE, EB are equal to the two CE, EF, each to each, in the triangles ABE, CFE; and the angle AEB is equal to the angle CEF, because they are opposite vertical angles; (1. 15.) therefore the base AB is equal to the base CF, (1. 4.) and the triangle AEВ to the triangle CEF, and the remaining angles of one triangle to the remaining angles of In the same manner, if the side BC be bisected, and AC be produced to G; it may be demonstrated that the angle BCG, that is, the angle ACD, (1. 15.) is greater than the angle ABC. Therefore, if one side of a triangle, &c. Q.E.D. Any two angles of a triangle are together less than two right angles. Let ABC be any triangle. Then any two of its angles together shall be less than two right angles. Produce any side BC to D. Then because ACD is the exterior angle of the triangle ABC; therefore the angle ACD is greater than the interior and opposite angle ABC; (1. 16.) to each of these unequals add the angle ACB; Therefore the angles ACD, ÂCB are greater than the angles ABC, ACB; but the angles ACD, ACB are equal to two right angles; (1. 13.) therefore the angles ABC, BCA are less than two right angles. In like manner it may be demonstrated, that the angles BAC, ACB are less than two right angles, Therefore any two angles of a triangle, &c. PROPOSITION XVIII. THEOREM. Q. E.D. The greater side of every triangle is opposite to the greater angle. Let ABC be a triangle, of which the side AC is greater than the side AB. Then the angle ABC shall be greater than the angle BCA. Since the side AC is greater than the side AB, make AD equal to AB, (1. 3.) and join BD. Then because AD is equal to AB in the triangle ABD, therefore the angle ADB is equal to the angle ABD, (1. 5.) but because the side CD of the triangle DBC is produced to A, therefore the exterior angle ADB is greater than the interior and opposite angle DCB; (1. 16.) but the angle ADB has been proved equal to the angle ABD, therefore the angle ABD is greater than the angle DCB; wherefore much more is the angle ABC greater than the angle ACB. Therefore the greater side, &c. Q.E.D. PROPOSITION XIX. THEOREM. 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. Then the side AC shall be greater than the side AB. A B For, if AC be not greater than AB, then the angle ABC would be equal to the angle ACB; (1.5.) but it is not equal; (hyp.) therefore the side AC is not equal to AB. then the angle ABC would be less than the angle ACB; (1. 18.) but it is not less, therefore the side AC is not less than AB; and AC has been shewn to be not equal to AB; therefore AC is greater than AB. Wherefore the greater angle, &c. Q. E.D. PROPOSITION XX. THEOREM. Any two sides of a triangle are together greater than the third side. Let ABC be a triangle. Then any two sides of it together shall be greater than the third side, viz. the sides BA, AC greater than the side BC; AB, BC greater than AC; and BC, CA greater than AB. B A D Produce the side BA to the point D, make AD equal to AC, (1. 3.) and join DC. Then because AD is equal to AC, therefore the angle ADC is equal to the angle ACD; (1. 5.) the angle BCD is greater than the angle BDC, (ax. 9.) and that the greater angle is subtended by the greater side; (1. 19.) therefore the side DB is greater than the side BC; but DB is equal to BA and AC, therefore the sides BÃ and AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA; also that BC, CA are greater than AB. Therefore any two sides, &c. Q.E.D. PROPOSITION XXI. THEOREM. If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle. Then BD and DC shall be less than BA and AC the other two sides of the triangle, but shall contain an angle BDC greater than the angle BAC. A E B Produce BD to meet the side AC in E. Because two sides of a triangle are greater than the third side, (1.20.) therefore the two sides BA, AE of the triangle ABE are greater than BE; to each of these unequals add EC; therefore the sides BA, AC are greater than BE, EC. (ax. 4.) Again, because the two sides CE, ED of the triangle CED are greater than DC; (1. 20.) add DB to each of these unequals ; therefore the sides CE, EB are greater than CD, DB. (ax. 4.) 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; (1. 16.) therefore the exterior angle BDC of the triangle CDE is greater than the interior and opposite angle CED; for the same reason, the exterior angle CED of the triangle ABE is greater than the interior and opposite angle BAC; and it has been demonstrated, that the angle BDC is greater than the angle CEB; much more therefore is the angle BDC greater than the angle BAC. Therefore, if from the ends of the side, &c. Q.E.D. |