PROPOSITION IX. PROBLEM 78. Construct the perpendicular-bisector of a given segment. E Given line segment AB. Required to construct the perpendicular-bisector of AB. Construction. 1. With A and B as centers, and with equal radii, draw arcs intersecting at C and also at D. 2. Statement. Draw CD intersecting AB at E. E bisects AB. Proof. 1. AC = BC, and also AD = BD. 2. .. CD is the perpendicular-bisector of AB. [If two points are each equidistant from the ends of a segment, they determine the perpendicular-bisector of the segment.] $ 17 $ 77 Ex. 65. Divide a given segment into four equal parts. Ex. 66 Draw a triangle of large size. Construct the perpendicularbisectors of the three sides. What happens? 79. A Median of a triangle is the line drawn from a vertex to the mid-point of the opposite side. Ex. 67. of the A. Draw a triangle of large size. Construct the three medians What happens? Ex. 68. Prove that the median drawn to the base of an isosceles triangle bisects the vertical angle. (Construct the figure.) Note. Supplementary Exercises 12-14, p. 274, can be studied now. PROPOSITION X. PROBLEM 80. At a point in a line, construct a perpendicular to the line. Given C, any point in line AB. B Required to construct a perpendicular to AB at C. Construction. 1. With C as center, and any radius, draw arcs intersecting AB at D and E respectively. 2. With D and E as centers and a radius greater than one half DE, draw two arcs intersecting at F. 4. .. CF is the perpendicular-bisector of DE. Why? [Since AB and DE are the same straight line.] 81. Proposition X proves that one perpendicular can be drawn to a line at a point in the line. It can be proved that only one perpendicular can be drawn to a line at a point in the line. For, if CP and DP were both perpendicular to AB at P, then 1 and 2 would both be right angles and hence would Ꮯ Ꭰ be equal. But 1 is greater than 2, for the whole is greater than any of its parts. Ex. 69. Prove CF perpendicular to DE (§ 80) by proving that ZFCDFCE, and then using § 26. PROPOSITION XI. PROBLEM 82. Construct a perpendicular to a line from a point not in the line. Given line AB and point C not in AB. Required to construct a to AB from C. Construction. 1. With C as center and a convenient radius, draw an arc intersecting AB at D and E respectively. 2. With D and E as centers, and equal radii, draw two arcs intersecting at F. 3. .. CF is the perpendicular-bisector of DE. Why ? [Since AB and DE are the same straight line.] Historical Note. This construction is attributed to Oenipodes of Chios (465 B.C.) 83. Proposition XI proves that one perpendicular can be drawn to a line from a point not in the line. It will be proved later (§ 88) that only one perpendicular can be drawn to a line from a point not in the line. It will also be proved (§ 164) that the perpendicular is the shortest segment from the point to the line. Note. Supplementary Exercises 15-16, p. 274, can be studied now. 84. The Distance from a point to a line is the length of the perpendicular from the point to the line. 85. An Altitude of a triangle is the perpendicular drawn from a vertex to the opposite side or the opposite side extended; as, AD. Ex. 70. How many altitudes does a triangle have? B D Ex. 71. Construct a triangle whose sides are 2 inches, 3 inches, and 4 inches, respectively. Construct the three altitudes of the triangle. Ex. 72. Construct an angle of 45°. (Use Prop. XI and Prop. VI.) Construct an angle of 135°; of 2210; of 6710. Note. The second and third designs below are drawn upon the first figure as a background. Can you discover how the first figure is constructed? Can you make an original design similar to these designs? 86. An Exterior Angle of a triangle is the angle at any vertex formed by a side of the triangle and the adjacent side extended; as, DCA. One interior angle is adjacent to the exterior B A C angle and the other two are remote interior angles. Thus, ZA and B are the remote interior angles of exterior LDCA. Ex. 73. Draw a large figure like that in § 86. ZDCA and each of the remote interior angles. angle compare with the remote interior angles ? Measure the exterior How does the exterior Ex. 74. In the figure for Prop. IX, p. 44, Z CEB is an exterior angle of what triangles ? 87. An exterior angle of a triangle is greater than either remote interior angle. PART I. Proof. 1. Through O, the mid-point of BC, draw AO. Extend 40 to E, making OE equal to AO. 2. 3. 4. .. ΔΑΒΟ ~ Δ 0CE. [Give the full proof.] <4: = Z1. ZBCD > Z1. [The whole is greater than any of its parts.] Draw CE. Why? Ax. 8, § 51 Ex. 75. In the figure for Prop. XII, prove that BOE is greater than 24. (Use § 87.) Ex. 76. Prove also that AOC > 21. Ex. 77. Compare ECF with 3. |