Construction. 1. Describe A1, A2, with radii rr' and r+r' respectively. centers O, and Figs. 1 and 2. Post. 6 2. From O'draw tangents O'C1, O'C2, to A1, A2. Pr. 3 3. Draw OC1, OC2, cutting circumference A at E1, E2. 4. Draw O'D1 || OE1, and O'D2 II E2O. Proof. 1. C1, C2 are rt. . 2. In Fig. 1, C1E1 = and II O'D1, ·.· C1E1=OE1—OС1=r—(r—r')=r'. 3... C1O'DE1 is a, and E1, D, are rt. . .. D1E, is tangent to 2 I, ths. 25, 23, cor. A, B. Th. 9, cor. 3 4. NOTE. In Fig. 3 the two circles have moved to external tangency, and the two interior tangents have closed up into one. In Fig. 4 the circumferences intersect and the interior tangents have vanished. In Fig. 5 the circles have become internally tangent and the two exterior tangents have closed up into one. In Fig. 6 the circle B lies wholly within the circle A, and the tangents have all vanished. EXERCISES. 337. All tangents drawn from points on the outer of two concentric circumferences to the inner are equal. 338. Find the locus of the centers of all circles touching two intersecting lines. (Show that it is a pair of perpendiculars.) Suppose the two lines were parallel instead of intersecting. 339. Describe a circle of given radius to touch two given lines. Show that a solution is, in general, impossible if the lines are parallel, but that otherwise there are four solutions. 340. From what two points in the plane are two circles seen under equal angles? Problem 5. On a given line-segment to construct a segment of a circle containing a given angle. 4. These will intersect YY' at the centers of the whose segments on AB are required Proof. 1. The two 1 from A, B, meet YY', as at O', O. I, th. 17, cor. 4 2. O is the center of O with chord AB and tangent BD. Th. 5, cor. 2; th. 9, cor. 4 3... ABD, or N, central on AĒB. 4. = 5. Similarly for segment Y'BA. Th. 13 Th. 11 EXERCISES. 341. In the figure of pr. 5, prove that AO' || OB. 342. Given an equilateral triangle, ABC, find a point P such that the circles circumscribing PBC, PCA, PAB are all equal. 343. To divide a circle into two segments so that the angle contained in one shall be double that contained in the other. 344. The bisectors of the interior and the exterior vertical angles of a triangle meet the circumscribed circumference in the mid-points of the arcs into which the base divides that circumference, and the line joining those points is the diameter which bisects the base. 345. A triangle whose angles are respectively 30°, 50°, 100° is inscribed in a circle; the bisectors of the angles meet the circumference in A, B, C. Find the number of degrees in the angles of ▲ ABC. APPENDIX TO BOOK III.- - METHODS. THE student has already been informed of three important methods of attacking a proposition, (1) by Analysis, (2) by Intersection of Loci (I, ths. 31, 32), and (3) by reductio ad absurdum. While these are the most valuable, and deserve further elaboration, there are others also worthy of mention. I. METHOD OF ANALYSIS. This method, first found in Euclid's Geometry, though attributed to Plato, may be thus described: Analysis is a kind of inverted solution; it assumes the proposition proved, considers what results follow, and continues to trace these results until a known proposition is reached. It then seeks to reverse the process and to give the usual or Synthetic proof. A more modern form of analysis is sometimes known as the Method of Successive Substitutions. In this the student substitutes in place of the given proposition another upon which the given one depends, and so on until a familiar one is reached. The student reasons somewhat as follows: "I can solve A if I can solve B, and I can solve B if I can solve C, but I can solve C"; or, "A is true if B is true, and B is true if C is true, but C is true, hence A and B are true.” EXERCISES. 346. Through a given point to draw a line to make equal angles with two intersecting lines. Analysis. Suppose x, y the lines, P the point, and the required line; then, in the figure, c = Za + <b; buta is to equal angle b, ../c=2a; .. if c is bisected, and a line is X drawn through P parallel to this bisector, the construction is effected. Now that the method is discovered, give the solution in the ordinary way. 347. Through a given point to draw a line such that the segments intercepted by the perpendiculars let fall upon it from two given points shall be equal. X D' Analysis. Suppose P the given point through which the line x is to be drawn, and A and B the other given points; then, in the figure, AD and BD' x, and DP is to equal PD'. Further, if AP is produced to meet BD' produced at A', then DPA ▲ D'PA', and.. AP PA'. But. A and P are given, AP can be drawn, and PA' found; .. A' can be found, and .. A'B; then from P a can be drawn to A'B, and the problem is solved. Always give the solution in the ordinary way. 348. If two circles are tangent, any secant drawn through their point of contact cuts off segments from one that contain angles equal to the angles in the segments of the other. Analysis. 1. Let CD be the common tangent to O, O' at their point of contact P. III, th. 20, cor. 2 2. Then an in segment A = an in segment A', if La= La. III, th. 13 Prel. th. 6 . A 349. Prove the theorem of ex. 348 when O' lies within OO and is tangent. 350. To inscribe a rhombus in a triangle, so as to have one of its angles coinciding with an angle of the triangle, and the opposite vertex of the rhombus lying on the opposite side of the triangle. 351. To draw a tangent to a given circle, perpendicular to a given line. 352. To construct a triangle, ABC, having given c, C, and the foot of the perpendicular from C to c. 353. From two given points to draw lines making equal angles with a given line, the points being on (1) the same side of the given line, (2) opposite sides of the given line. 354. To draw, through a given point, a secant from which equal circumferences shall cut off equal chords. Discuss the number of solutions for various positions of the given point. 355. Through one of the points of intersection of two circumferences draw a chord of one circle which shall be bisected by the circumference of the other. II. METHOD OF INTERSECTION OF LOCI. This method, adapted chiefly to the solution of problems, has already been used in Book I (ths. 31, 32). So long as it is known merely that a point is on one line, its position is not definitely known; but if it is known that it is also on another line, its position may be uniquely determined. For example, if it is known that a point is on each of two intersecting lines, it is uniquely determined as their point of intersection; but if it is on a straight line and a circumference which the line intersects, it may be either of the two points of intersection. For convenience of reference the following theorems are stated, and will be referred to by the letters prefixed : a. The locus of points at a given distance from a given point is the circumference described about that point as a center, with a radius equal to the given distance. (I, th. 28.) b. The locus of points at a given distance from a given line consists of a pair of parallels at that distance, one on each side of the fixed line. (I, th. 30, cor. 2.) c. The locus of points equidistant from two given points is the perpendicular bisector of the line joining them. (I, th. 29.) d. The locus of points equidistant from two given lines consists of the bisectors of their included angles; if the lines are parallel, it consists of a parallel midway between them. (I, th. 30, and cor. 1.) e. The locus of points from which a given line subtends a given angle is an arc subtended by that chord. (III, th. 11, cor. 3.) ABBREVIATIONS. In this appendix the following abbreviations will be used: In the triangle ABC the sides will, as usual, be designated by a, b, c; the altitudes on those sides by ha, hb, h, respectively; the corresponding medians by ma, mb, mc; the corresponding angle-bisectors of A, B, C, terminated by a, b, c, by Va, V, Vc; the radii of the inscribed |