Demonstration. Then, since the straight line AB, standing on the straight line CD, makes the adjacent angles ABC, ABD equal to one another, each of them is a right angle (Def. 10); therefore 1. The angles ABC, ABD are two right angles. Secondly, let them not be equal to one another. B Construction. From the point B draw BE at right angles to CD (I. 11). Demonstration. Because BE is at right angles to CD (constr.), therefore 1. The angles CBE, EBD are two right angles (Def. 10). And because the angle CBE is equal to the angles CBA, ABE, add the angle EBD to each of these equals, therefore 2. The angles CBE, EBD are equal to the three angles CBA, ABE, EBD (Ax. 2). Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC. Therefore 3. The angles DBA, ABC are equal to the three angles DBE, EBA, ABC. But the angles CBE, EDD, have been proved equal to the same three angles, and things which are equal to the same thing are equal to one another. Therefore EBD 4. The angles CBE, ÉDD are equal to the angles DBA, ABC, but the angles CBE, EBD are two right angles, therefore 5. The angles DBA, ABC are together equal to two right angles (Ax. 1). Wherefore, the angles which one straight line, &c. Q.E.D. PROPOSITION 14.-Theorem. 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, then 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 BC. Construction. For, if BD be not in the same straight line with BC; if possible, let BE be in the same straight line with it. Demonstration. Then because AB meets the straight line CBE, therefore 1. The adjacent angles CBA, ABE are equal to two right angles (I. 13); but the angles CBA, ABD, are equal to two right angles (hyp.), therefore 2. The angles CBA, ABE are equal to the angles CBA, ABD (Ax. 1); take away from these equals the common angle CBA, therefore 3. The remaining angle ABE is equal to the remaining angle ABD (Ax. 3), the less angle equal to the greater, which is impossible ; therefore 4. BE is not in the same straight line with BC. And in the same manner it may be demonstrated, that no other can be in the same straight line with it but BD, therefore 5. BD is in the same straight line with BC. Wherefore, if at a point, &c. Q.E.D. 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 3. The angles CEA, AED are equal to the angles AED, DEB (Ax. 1); take away from each the common angle AED, and 4. The remaining angle CEA is equal to the remaining angle DEB (Ax. 3). 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 E B 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.), 1. The two sides AE, EB are equal to the two CE, EF, each to each, in the triangles ABE, CFE; and 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 2. The angles ACD, ACB are greater than the angles ABC, ACB, but 3. The angles ACD, ACB are equal to two right angles (I. 13) ; therefore 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. 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 A B (hyp.), make AD equal to AB (I. 3) and join BD. |