PROPOSITION 15.-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. Demonstration. Because the straight line AE makes with CD, at the point E, the adjacent angles CEA, AED. 1. The angles CEA, AED are together equal to two right angles (I. 13). Again, because the straight line DE makes with AB, at the point E, the adjacent angles AED, DEB, 2. The angles AED, DEB are also equal to two right angles (I. 13), but the angles CEA, AED have been shown to be equal to two right angles, therefore take away 3. The angles CEA, AED are equal to the angles AED, DEB (Ax. 1); from each the common angle AED, and 4. The remaining angle CEA is equal to the remaining angle DEB (Ax. 3). In the same manner it may be demonstrated that 5. The angle CEB is equal to the angle AED. Therefore, if two straight lines cut one another, &c. Q.E.D. 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. PROPOSITION 16.-Theorem. 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 the side BC be produced to D. Then the exterior angle ACD shall be greater than either of the interior opposite angles CBA or BAC. A B G Construction. Bisect AC in E (I. 10), and join BE; produce BE to F, making EF equal to BE (I. 3) and join FC. Demonstration. Because AE is equal to EC, and BE to EF (constr.), and 1. The two sides AE, EB are equal to the two CE, EF, each to each, in the triangles ABE, CFE; 2. The angle AEB is equal to the angle CEF, because they are opposite, vertical angles (I. 15), therefore 3. The base AB is equal to the base CF, and the triangle A EB to the triangle CEF (I. 4), and the remaining angles of one triangle to the remaining angles of the other, each to each, to which the equal sides are opposite. Wherefore 4. The angle BAE is equal to the angle ECF, but the angle ECD or ACD is greater than the angle ECF, therefore 5. The angle ACD is greater than the angle BAE or BAC. 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 (I. 15) is greater than the angle ABC. Therefore, if one side of a triangle, &c. Q.E.D. PROPOSITION 17.-Theorem. 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. Construction. Produce any side BC to D. Demonstration. Then because ACD is the exterior angle of the triangle ABC, therefore 1. The angle ACD is greater than the interior and opposite angle ABC (I. 16); to each of these unequals add the angle ACB, therefore but therefore 2. The angles ACD, ACB are greater than the angles ABC, ACB, 3. The angles ACD, ACB are equal to two right angles (I. 13) ; 4. The angles ABC, ACB are less than two right angles. In like manner it may be demonstrated that 5. The angles BAC, ACB, as also CAB, ABC are less than two right angles. Therefore any two angles of a triangle, &c. Q.E.D. PROPOSITION 18.-Theorem. 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 ACB. Construction. Since the side AC is greater than the side AB (hyp.), make AD equal to AB (I. 3) and join BD. Demonstration. Then, because AD is equal to AB, in the triangle ABD, therefore D B 1. The angle ABD is equal to the angle ADB, (I. 5), but because the side CD of the triangle BDC is produced to A, therefore 2. The exterior angle ADB is greater than the interior and opposite angle DCB (I. 16), but the angle ADB has been proved equal to the angle ABD, therefore 3. The angle ABD is greater than the angle DCB; wherefore much more 4. The angle ABC is greater than the angle ACB. Therefore the greater side, &c. Q.E.D. PROPOSITION 19.-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. B For, if AC be not greater than AB, AC must either be equal to or less than AB; if AC were equal to AB, then the angle ABC would be equal to the angle ACB (Î. 5), but it is not equal (hyp.), therefore 1. The side AC is not equal to AB. Again, if AC were less than AB, then the angle ABC would be less than the angle ACB (I. 18), but it is not less (hyp.), therefore 2. The side AC is not less than AB, and AC has been shown to be not equal to AB, therefore 3. AC is greater than AB. Wherefore the greater angle, &c. Q.E.D. PROPOSITION 20.-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 sideviz., the sides BA, AC greater than the side BC; AB, BC greater than AC; and BC, CA greater than AB. B Construction. Produce the side BA to the point D, make AD equal to AC (I. 3), and join DC. Demonstration. Then because AD is equal to AC (constr.), therefore 1. The angle ACD is equal to the angle ADC (I. 5), but the angle BCD is greater than the angle ACD (Ax. 9), therefore also 2. The angle BCD is greater than the angle ADC. And because in the triangle DBC, the angle BCD is greater than the angle BDC, and that the greater angle is subtended by the greater side (I. 19), therefore 3. The side DB is greater than the side BC, but DB is equal to BA and AC, therefore 4. The sides BA, AC are greater than BC. In the same manner it may be demonstrated, that 5. The sides AB, BC are greater than CA; and BC, CA are greater than AB. Therefore any two sides, &c. Q.E.D. |