PROPOSITION XXXIX. PROBLEM 250. To divide a given straight line into any number of equal parts. K H B Given straight line AB. To divide AB into n equal parts. I. Construction 1. Draw the unlimited line AX. 2. Take any convenient segment, as AR, and, beginning at A, lay it off n times on AX, 3. Connect the nth point of division, as K, with B. 4. Through the preceding point of division, as P, draw a line PH KB. § 188. 5. Then HB is one nth of AB. 6. .. HB, if laid off successively on AB, will divide AB into n equal parts. II. The proof and discussion are left as an exercise for the student. Ex. 305. Divide a straight line into 7 equal parts. Ex. 306. eter. Construct an equilateral triangle, having given the perim Ex. 307. Construct a square, having given the perimeter. 251. Def. The line joining the mid-points of the nonparallel sides of a trapezoid is called the median of the trapezoid. Ex. 308. Show, by generalizing, that Cor. III, Prop. XXXVIII, may be obtained from Cor. I and Cor. IV from Cor. II, Ex. 309. The lines joining the mid-points of the sides of a quadrilateral taken in order form a parallelogram. Ex. 310. What additional statement can you make if the quadrilateral in Ex. 309 is an isosceles trapezoid? a rectangle? a rhombus ? a square? Ex. 311. Lines drawn from the mid-point of the base of an isosceles triangle to the mid-points of its equal sides form a rhombus or a square. When is the figure a rhombus ? when a square? Ex. 312. The mid-points of the sides of a quadrilateral and the midpoints of its two diagonals are the vertices of three parallelograms whose diagonals are concurrent. Ex. 313. What is the perimeter of each parallelogram in Ex. 312? Ex. 315. Through a given point within an angle construct a line, limited by the two sides of the angle, and bisected at the given point. Ex. 316. Every diagonal of a parallelogram is trisected by the lines joining the other two vertices with the mid-points of the opposite sides. Ex. 317. If a triangle inscribed in another triangle has its sides parallel respectively to the sides of the latter, its vertices are the midpoints of the sides of the latter. Ex. 318. If the lower base KT of trapezoid RSTK is double the upper base RS, and the diagonals intersect at O, prove OK double OS, and QT double OR. Ex. 319. Construct a trapezoid, given the two bases, one diagonal, and one of the non-parallel sides. In the two following exercises prove the properties which require proof, state those which follow by definition, and those which have been proved in the text: Ex. 320. Properties possessed by all trapezoids : (a) Two sides of a trapezoid are parallel. (b) The two angles adjacent to either of the non-parallel sides are supplementary. (c) The median of a trapezoid is parallel to the bases and equal to one half their sum. Ex. 321. In an isosceles trapezoid: (a) The two non-parallel sides are equal. (b) The angles at each base are equal and the opposite angles are supplementary. (c) The diagonals are equal, PROPOSITION XL. THEOREM 252. The two perpendiculars to the sides of an angle from any point in its bisector are equal. Given ABC; P any point in BR, the bisector of ZABC; PD and PE, the Is from P to BA and BC respectively. To prove PD = PE. ARGUMENT ONLY 1. In rt. A DBP and PBE, PB = PB. 2. Z DBP = LPBE. 3. .. ΔΟΒΡ=ΔΡΒΕ. 4. .. PD PE. 253. Prop. XL may be stated as follows: Q.E.D. Every point in the bisector of an angle is equidistant from the sides of the angle. Ex. 322. Find a point in one side of a triangle which is equidistant from the other two sides of the triangle. Ex. 323. Find a point equidistant from two given intersecting lines and also at a given distance from a fixed third line. Ex. 324. Find a point equidistant from two given intersecting lines and also equidistant from two given parallel lines. Ex. 325. Find a point equidistant from the four sides of a rhombus. Ex. 326. The two altitudes of a rhombus are equal. Prove. Ex. 327. Construct the locus of the center of a circle of given radius, which rolls so that it always touches the sides of a given angle. Do not PROPOSITION XLI. THEOREM (Opposite of Prop. XL) 254. The two perpendiculars to the sides of an angle from any point not in its bisector are unequal. Given ABC; P any point not in BR, the bisector of ABC; PD and PE, is from P to BA and BC respectively. To prove PD‡PE. OUTLINE OF PROOF Draw FG BA; draw PG. Then FEFG. Now PF+ FG > PG. .. PE> PG. But PG> PD. .. PE > PD. 255. Prop. XLI may be stated as follows: Every point not in the bisector of an angle is not equidistant from the sides of the angle. 256. Cor. I. (Converse of Prop. XL). Every point equidistant from the sides of an angle lies in the bisector of the angle. HINT. Prove directly, using the figure for § 252, or apply § 137. 257. Cor. II. The bisector of an angle is the locus of all points equidistant from the sides of the angle. Ex. 328. What is the locus of all points that are equidistant from a pair of intersecting lines? CONCURRENT LINE THEOREMS PROPOSITION XLII. THEOREM 258. The bisectors of the angles of a triangle are concurrent in a point which is equidistant from the three sides of the triangle. B Given ABC with AF, BE, CD the bisectors of A, B, and C respectively. To prove: (a) AF, BE, CD concurrent in some point as 0; (b) the point o equidistant from AB, BC, and CA. |