PROPOSITION V. PROBLEM 53. To erect a perpendicular to a line at a given point on that line. Let AB be the given line, and C the given point on the line. Required to erect a perpendicular to AB at C. Lay off CD CE. With D and E as centers, and with a radius greater than DC (one half of DE), describe two arcs intersecting at F. Join F and C. FC is the required perpendicular. For, F and c are each equally distant from D and E (construction). .. by § 49, FC is perpendicular to DE or AB. Q.E.F. 54. EXERCISE. To construct a R.A. A, having given the two sides about the R.A. ▲ CDE is the required ▲ because it fulfills all the required conditions; i.e. it is right angled at C, and the sides about C are equal respectively to m and n. Q.E.F PROPOSITION VI. PROBLEM 55. To bisect a given line. B M Let AB be the given line. Required to bisect it. With A and B as centers, and with any radius greater than one half of AB, describe arcs intersecting at C and D. Draw CD. Then will CD bisect AB. For, the points C and D are each equally distant from the extremities of AB (construction). .. CD bisects AB (§ 49). 56. EXERCISE. Divide a given line into quarters. Q.E.F. 57. EXERCISE. If the radius used for describing the two arcs that intersect at C in the figure of Prop. VI is greater than the radius used for describing the two arcs that intersect at D, will CD bisect AB? 58. EXERCISE. When will the lines AB and CD bisect each other? 59. EXERCISE. In a given line find a point that is equally distant from two given points. When is this problem impossible? PROPOSITION VII. THEOREM 60. The sum of the adjacent angles formed by one line meeting another, is two right angles. Proof. Erect BE perpendicular to CD at B. EBC and EBD are R. A.'s. LABC=1 R.A. + EBA. ZABD = 1 R.A. — ▲ EBA. (§ 53.) (1) (2) Adding (1) and (2), ▲ ABC + Z ABD = 2 R.A's. Q.E.D. 61. COROLLARY I. If one of two adjacent angles formed by one line meeting another is a right angle, the other is also a right angle. 62. COROLLARY II. If two straight lines intersect each other, and one of the angles formed is a right angle, the other three angles are also right angles. 63. COROLLARY III. The sum of all the angles formed at a point in a line, and on the same side of the line, is two right angles. SUGGESTION. Show that the sum of all the angles at C equals E F ZFCA+ZFCB or LGCA + ZGCB, etc. A B Adjacent angles formed by one line meeting another are supplementary adjacent angles. 66. DEFINITION. If two angles are together equal to one right angle, they are called complementary angles. Each angle is the complement of the other. 67. EXERCISE. Find the supplement and also the complement of each of the following angles: R.A., ↓ R.A., ¦ R.A. Find the value of each of two supplementary angles, if one is five times the other. 68. EXERCISE. complement. Given an angle, construct its supplement and also its 71. EXERCISE. Find the supplement of the complement of also the complement of the supplement of 14 R.A. R.A., 72. DEFINITION. One proposition is the converse of another, when the hypothesis and conclusion of one are respectively the conclusion and hypothesis of the other. The converse of a proposition is not necessarily true. We shall prove later (see § 85) that "if the sides of one triangle are equal respectively to the sides of another, the angles of the first triangle are equal respectively to those of the second." Show, by drawing triangles, that the converse of this proposition, i.e. "if the angles of one triangle are equal respectively to the angles of another, the sides of the first triangle are equal respectively to those of the second," is not necessarily true. PROPOSITION VIII. THEOREM (CONVERSE OF PROP. VII.) 73. If the sum of two adjacent angles is two right angles, their exterior sides form a straight line. C Let D CDA+CDB = 2 R.A.'s. To Prove AD and DB form a straight line. Proof. Suppose DB is not the prolongation of AD, and that some other line, as DE, is. By § 60 CDA + ≤ CDE would equal 2 R.A.'s. = By Axiom 1., ≤ CDA + ≤ CDE would equal ≤ CDA + ≤ CDB. Whence CDE would equal ≤ CDB. This contradicts Axiom 11. Therefore the supposition that DB is not the prolongation of AD is false, and AD and DB form a straight line. Q.E.D. |