also into two unequal parts, the rectangle of the unequal parts with the square on the line between the points of section, is equal to the square on half the line. (Euclid, B. II, prop. 5.) 49. The square on the straight line, drawn from the vertex of an isosceles triangle to any point in the base, is less than the square on a side of the triangle by the rectangle of the segments of the base. 50. The sides AB, CD of a quadrilateral ABCD inscribed in a circle are produced to meet in P; and PE, PF are drawn perpendicular to AD, BC respectively: prove that AE is to ED as CF to FB. 51. The sides AB, AD, produced if necessary, of a parallelogram ABCD, meet a line through C in E, F respectively; CB, CD, produced if necessary, meet a line through A in G, H respectively. Prove that GE is parallel to HF. PROBLEMS. 52. Construct a square equal to three given squares. 53. Construct a square which shall be five times as great as a given square. 54. Construct a triangle equal in area to a given quadrilateral. 55. Construct an isosceles triangle equal in area to a given triangle and having a given vertical angle. 56. Divide a straight line into two parts, so that the sum of the squares on the parts may be equal to a given square. 57. Trisect a triangle by straight lines drawn from a given point in one of its sides. 58. Find a point O inside a triangle ABC such that the triangles OAB, OBC, OCA are equal. 59. Construct a square that shall be one-third of a given square. 60. Divide a straight line into two parts so that the rectangle contained by the whole and one part may be equal to a given square. Let AB be the given line. Draw BD to AB so that BD = given square; join AD, draw DE 1 to AD, meeting AB produced in E. 61. Produce AB to C so that the rectangle of AB and AC may be equal to a given square. 62. Construct a parallelogram equal to a given triangle and having one of its angles equal to a given angle. 63. Given three similar triangles: construct another similar triangle and equivalent to their sum. (377.) 64. Construct a triangle similar to a given triangle ABC which shall be to ABC in the ratio of AB to BC. Find DE a mean proportional between AB, BC, etc. 65. Bisect a triangle by a straight line parallel to one of its sides. Bisect AC of the ▲ ABC in D; from AC cut off AE a mean proportional to AC, AD, etc. 66. A is a point on a given circle: draw through A a straight line PAQ meeting the circle in P and a given straight line in Q, so that the ratio of PA to AQ may be equal to a given ratio. Draw AB to any pt. B in the given line MN; produce BA to C so that BA: AC given ratio, etc. 67. From the vertex of a triangle draw a line to the base, so that it may be a mean proportional between the segments of the base. About the given ▲ describe a O, etc. 68. Show how to draw through a given point in a side of a triangle a straight line dividing the triangle in a given ratio. 69. Construct a triangle equal to a given triangle and having one of its angles equal to an angle of the triangle, and the sides containing this angle in a given ratio. 70. Draw through a given point a straight line, so that the part of it intercepted between a given straight line and a given circle may be divided at the given point in a given ratio. Let P be the given pt., XY the given line, A the cent. of given : produce AP to meet XY in Q; in PA take B so that QP: PB = given ratio, etc. Book V. REGULAR POLYGONS. THE CIRCLE. REGULAR POLYGONS. 399. DEF. A regular polygon is a polygon which is both equilateral and equiangular; as, for example, the equilateral triangle and the square. Proposition 1. Theorem. 400. If the circumference of a circle be divided into any number of equal arcs, (1) the chords of these arcs form a regular polygon inscribed in the circle, and (2) the tangents at the points of division form a regular polygon circumscribed about the circle. Hyp. Let the Oce be divided into equal arcs at the pts. A, B, C, etc.; let AB, BC, CD, etc., be chords of these arcs, and GBH, HCK, etc., be F tangents. (1) To prove that ABCD regular polygon. .... is a L H Proof. Since arc AB =arc BC= arc CD = etc., (Hyp.) ... chd. AB chd. BC = chd. CD = and in the same etc., equal arcs are subtended by equal chords (194), (2) To prove that GHK... is a regular polygon. AB = BC = CD= etc., and GAB GBA = 2 HBC = /HCB = etc., being measured by halves of equal arcs (243). (399) ..the As ABG, BCH, etc., are all equal and isosceles. M (399) Q. E.D. 401. COR. 1. If a regular inscribed polygon is given, the tangents at the vertices of the given polygon form a regular circumscribed polygon of the same number of sides. 402. COR. 2. If a regular inscribed polygon ABCD.... is given, the tangents at the middle points M, N, P, etc., of the arcs AB, BC, CD, etc., form a regular circumscribed polygon whose sides are parallel to those of the inscribed polygon, and whose vertices A', B', C', etc., lie on O N the radii OAA', OBB', etc. For, the sides AB, A'B' are ||, being to OM, (204) and (210), and the same for the others; also, since B'M = B'N (268), the pt. B' must lie on the bisector OB (160) of the MON. 403. COR. 3. If the chords AM, MB, BN, etc., be drawn, the chords form a regular inscribed polygon of double the number of sides of ABCD.... 404. COR. 4. If through the points A, B, C, etc., tangents are drawn intersecting the tangents A'B', B'C', etc., a regular circumscribed polygon is formed of double the number of sides of A'B'C'D'.... 405. SCH. It is clear that the area of any inscribed polygon is less than the area of the inscribed polygon of double the number of sides; and the area of a circumscribed polygon is greater than that of the circumscribed polygon of double the number of sides. Proposition 2. Theorem. 406. A circle may be circumscribed about a regular polygon, and a circle may be inscribed in it. Hyp. Let ABCDEF be a regular polygon. (1) To prove that a O may be circumscribed about it. Proof. Bisect the s A and B, and let the bisectors meet at 0. In the same way it may be shown that each of the |