1. At the point B in the straight line AB, let the two straight lines BC, BD, on the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles: BD shall be in the same straight line with CB. Then because the 3. straight line AB makes with the straight line CBE, on one side of it, the angles ABC, ABE, B these angles are together equal to two right angles. [I. 13.] But the angles ABC, ABD are also together equal to two right angles. [Hyp.] Therefore the angles ABC, ABE are equal to the angles ABC, ABD. From each of these equals take away the common angle ABC, and the remaining angle ABE is equal to the remaining angle ABD, [Ax. 3.] the less to the greater; which is impossible. Therefore BE is not in the same straight line with CB. And in like manner it may be proved that no other can be in the same straight line with it but BD; therefore BD is in the same straight line with CB. Therefore, if at a point, &c. Q.E.D. It is important to notice that CB and BD must lie on opposite sides of BA, otherwise they will not necessarily be in the same straight line. Thus the lines BC, BD in the accompanying fig. make with AB the angles ABC, ABD equal to two right angles, but they are not on opposite sides of AB, and are therefore not in the same straight line. Ex. Show that four equal squares may be placed so as to form one square. IA PROPOSITION 15. THEOREM. If two straight lines cut one another, the vertical, or opposite, angles shall be equal. Let the two 1. shall be equal to the angle DEB, and the angle CEB to the angle AED. A E B 3. Because the straight line AE makes with the straight line CD the angles CEA, AED, these angles are together equal to two right angles. [I. 13.] Again, because the straight line DE makes with the straight line AB the angles AED, DEB, these also are together equal to two right angles. [I. 13.] But the angles CEA, AED have been proved to be together equal to two right angles. Therefore the angles CEA, AED are equal to the angles AED, DEB. From each of these equals take away the common angle AED, and the remaining angle CEA is equal to the remaining angle DEB. [Ax. 3.] In the same manner it can be proved that the angle CEB is equal to the angle AED. Therefore, if two straight lines, &c. Q.E.D. Corollary 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. Corollary 2. And consequently, that all the angles made by any number of lines meeting in one point, are together equal to four right angles. The two corollaries should be proved. Ex. 1. If the angle CEA in the above figure is a right angle, all the angles at E are right angles. 2. Let the four angles at E, taken in order, be bisected by the four straight lines EG, EH, EK, and EL. Show, 1. That GE and KE are in the same straight line, and also that HE and LE are in the same straight line. 2. That GK and HL are at right angles to one another. PROPOSITION 16. THEOREM. If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles. Let ABC be a triangle, and let one side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC. 1. Bisect AC in E, [I. 10.] join BE and produce it 2. to F, making EF equal to EB, [I. 3.] and join FC. 3. Because AE is equal to EC, and BE to EF; [Const.] the two sides AE, EB are equal to the two sides CE, EF, each to each; and the angle AEB is equal to the angle CEF, because they are opposite vertical angles; [I. 15.] therefore the base AB is equal to the base CF, and the triangle AEB to the triangle CEF, B and the remaining angles to the remaining angles, each to each, to E D which the equal sides are opposite; [I. 4.] therefore the angle BAE is equal to the angle ECF. But the angle ECD is greater than the angle ECF. [Ax. 9.] Therefore the angle ACD is greater than the angle BAE. In the same manner if BC be bisected, and the side AC be produced to G, it may be proved that the angle BCG, that is the angle ACD, is greater than the angle ABC. [I. 15.] Therefore, if one side, &c. Q.E.D. Ex. Construct the figure mentioned at the end of the proposition, and show that ACD is greater than ABC. 2. D. PROPOSITION 17. THEOREM. Any two angles of a triangle are together less than two right angles. 1. Let ABC be any triangle: any two of its angles are together less than two right angles. Produce BC to Then because 3. ACD is the exterior angle of the triangle ABC, it is greater than the interior and opposite angle ABC. [I. 16.] To each of these add the angle ACB. Therefore the angles ACD, ACB are greater than the angles ABC, ACB. But the angles ACD, ACB are together equal to two right angles. [I. 13.] Therefore the angles ABC, ACB are together less than two right angles. In like manner it may be proved that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles. Therefore, any two angles, &c. Q.E.D. Ex. 1. Prove this proposition by joining A to any point E in BC. 2. Only one perpendicular can be drawn from the same point to the same straight line. 3. The equal angles of an isosceles triangle must be both acute. 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: the angle ABC is also greater than the angle ACB. of the triangle BDC, it is greater than the interior and opposite angle DCB. [I. 16.] But the angle ADB is equal to the angle ABD [I. 5.], because the side AD is equal to the side AB. [Const.] Therefore the angle ABD is also greater than the angle ACB. Much more then is the angle ABC greater than the angle ACB. [Ax. 9.] Therefore, the greater side, &c. Q.E.D. A This proposition should be compared with prop. 5. Placed side by side they stand thus: Prop. 5. If two sides of a triangle are equal, the angles opposite to them are also equal. Prop. 18. If two sides of a triangle are unequal, the angles opposite to them are unequal; and the greater side has the greater angle opposite to it. A similar comparison may be made between the 6th and 19th. Ex. Prove this proposition by producing the less side AB to E, making AE equal to AC, and joining CE. PROPOSITION 19. THEOREM. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. 1. Let ABC be a triangle, of which the angle ABC is greater than the angle ACB: the side AC is also greater than the side AB. A For if not, AC 3. is either equal to AB or less than AB. |