59. If the three points be such as when joined by straight lines a triangle is formed; the points at which the inscribed circle touches the sides of the triangle, are the points at which the three circles touch one another. Euc. IV. 4. Different cases arise from the relative position of the three points. 60. Bisect the angle contained by the two lines at the point where the bisecting line meets the circumference, draw a tangent to the circle and produce the two straight lines to meet it. In this triangle inscribe a circle. 61. From the given angle draw a line through the center of the circle, and at the point where the line intersects the circumference, draw a tangent to the circle, meeting two sides of the triangle. The circle inscribed within this triangle will be the circle required. 62. Let the diagonal AD cut the arc in P, and let O be the center of the inscribed circle. Draw OQ perpendicular to AB. Draw PE a tangent at P meeting AB produced in E: then BE is equal to PD. Join PQ, PB. Then AB may be proved equal to QE. Hence AQ is equal to BE or DP. 63. Suppose the center of the required circle to be found, let fall two perpendiculars from this point upon the radii of the quadrant, and join the center of the circle with the center of the quadrant and produce the line to meet the arc of the quadrant. If three tangents be drawn at the three points thus determined in the two semicircles and the arc of the quadrant, they form a right-angled triangle which circumscribes the required circle. 64. Let AB be the base of the given segment, C its middle point. Let DCE be the required triangle having the sum of the base DE and perpendicular CF equal to the given line. Produce CF to H making FH equal to DE. Join HD and produce it, if necessary, to meet AB produced in K. Then CK is double of DF. Draw DL perpendicular to CK. 65. From the vertex of the isosceles triangle let fall a perpendicular on the base. Then, in each of the triangles so formed, inscribe a circle, Euc. IV. 4; next inscribe a circle so as to touch the two circles and the two equal sides of the triangle. This gives one solution: the problem is indeterminate. 66. If BD be shewn to subtend an arc of the larger circle equal to one-tenth of the whole circumference :-then BD is a side of the decagon in the larger circle. And if the triangle ABD can be shewn to be inscriptible in the smaller circle, BD will be the side of the inscribed pentagon. 67. It may be shewn that the angles ABF, BFD stand on two arcs, one of which is three times as large as the other. 68. It may be proved that the diagonals bisect the angles of the pentagon, and the five-sided figure formed by their intersection, may be shewn to be both equiangular and equilateral. 69. The figure ABCDE is an irregular pentagon inscribed in a circle; it may be shewn that the five angles at the circumference stand upon arcs whose sum is equal to the whole circumference of the circle; Euc. III. 20. 70. If a side UD (figure, Euc. IV. 11) of a regular pentagon be produced to K, the exterior angle ADK of the inscribed quadrilateral figure ABCD is equal to the angle ABC, one of the interior angles of the pentagon. From this a construction may be made for the method of folding the ribbon. 71. In the figure, Euc. iv. 10, let DC be produced to meet the circumference in F, and join FB. Then FB is the side of a regular pentagon inscribed in the larger circle, D is the middle of the arc subtended by the adjacent side of the pentagon. Then the difference of FD and BĎ is equal to the radius AB. Next, it may be shewn, that FD is divided in the same manner in C as AB, and by Euc. II. 4, 11, the squares on FD and DB are three times the square on AB, and the rectangle of FD and DB is equal to the square on AB. 72. If one of the diagonals be drawn, this line with three sides of the pentagon forms a quadrilateral figure of which three consecutive sides are equal. The problem is reduced to the inscription of a quadrilateral in a square. 73. This may be deduced from Euc. Iv. 11. 74. The angle at A the center of the circle (fig. Euc. Iv. 10) is onetenth of four right angles, the arc BD is therefore one-tenth of the circumference, and the chord BD is the side of a regular decagon inscribed in the larger circle. Produce DC to meet the circumference in F and join BF, then BF is the side of the inscribed pentagon, and AB is the side of the inscribed hexagon. Join FA. Then FCA may be proved to be an isosceles triangle and FB is a line drawn from the vertex meeting the base produced. If a perpendicular be drawn from F on BC, the difference of the squares on FB, FC may be shewn to be equal to the rectangle AB, BC, (Euc. 1. 47; 11. 5. Cor.); or the square on AC. 75. Divide the circle into three equal sectors, and draw tangents to the middle points of the arcs, the problem is then reduced to the inscription of a circle in a triangle. 76. Let the inscribed circles whose centers are A, B touch each other in G, and the circle whose center is C, in the points D, E; join A, D; A, E; at D, draw DF perpendicular to DA, and EF to EB, meeting in F. Let F, G be joined, and FG be proved to touch the two circles in G whose centers are A and B. 77. The problem is the same as to find how many equal circles may be placed round a circle of the same radius, touching this circle and each other. The number is six. 78. This is obvious from Euc. IV. 7, the side of a square circumscribing a circle being equal to the diameter of the circle. 79. Each of the vertical angles of the triangles so formed, may be. proved to be equal to the difference between the exterior and interior angle of the heptagon. 80. Every regular polygon can be divided into equal isosceles triangles by drawing lines from the center of the inscribed or circumscribed circle to the angular points of the figure, and the number of triangles, will be equal to the number of sides of the polygon. If a perpendicular FG be let fall from F (figure, Euc. IV. 14) the center on the base CD of FCD, one of these triangles, and if GF be produced to H till FH be equal to FG, and HC, HD be joined, an isosceles triangle is formed, such that the angle at H is half the angle at F. Bisect HC, HD in K, L, and join KL; then the triangle HKL may be placed round the vertex H, twice as many times as the triangle CFD round the vertex F. 81. The sum of the arcs on which stand the 1st, 3rd, 5th, &c. angles, is equal to the sum of the arcs on which stand the 2nd, 4th, 6th, &c. angles. 82. The proof of this property depends on the fact, that an isosceles triangle has a greater area than any scalene triangle of the same perimeter. GEOMETRICAL EXERCISES ON BOOK VI. HINTS, &c. 6. In the figure Euc. vi. 23, let the parallelograms be supposed to be rectangular. Then the rectangle AC: the rectangle DG :: BC: CG, Euc. vi. 1. and the rectangle DG: the rectangle CF :: CD: EC, whence the rectangle AC: the rectangle CF :: BC. CD: CG. EC. In a similar way it may be shewn that the ratio of any two parallelograms is as the ratio compounded of the ratios of their bases and altitudes. 7. Let two sides intersect in O, through O draw POQ parallel to the base AB. Then by similar triangles, PO may be proved equal to OQ and POFA, QOEB, are parallelograms: whence AE is equal to FB. 8. Apply Euc. vi. 4, v. 7. 9. Let ABC be a scalene triangle, having the vertical angle A, and suppose ADE an equivalent isosceles triangle, of which the side AD is equal to AE. Then Euc. vi. 15, 16, AC.AB AD.AE, or AD'. Hence AD is a mean proportional between AC, AB. Euc. vI. 8. 10. The lines drawn making equal angles with homologous sides, divide the triangles into two corresponding pairs of equiangular triangles; by Euc. vi. 4, the proportions are evident. 11. By constructing the figure, the angles of the two triangles may easily be shewn to be respectively equal. 12. A circle may be described about the four-sided figure ABDC. By Euc. 1. 13; Euc. I. 21, 22. The triangles ABC, ĂCE may be shewn to be equiangular. 13. Apply Euc. 1. 48; 11. 5. Cor., vi. 16. 14. This property follows as a corollary to Euc. vi. 23: for the two triangles are respectively the halves of the parallelograms, and are therefore in the ratio compounded of the ratios of the sides which contain the same or equal angles: and this ratio is the same as the ratio of the rectangles by the sides. 15. Let ABC be the given triangle, and let the line EGF cut the base BC in G. Join AG. Then by Euc. vI. 1, and the preceding theorem (14) it may be proved that AC is to AB as GE is to GF. 16. The two means and the two extremes form an arithmetic series of four lines whose successive differences are equal; the difference therefore between the first and the fourth, or the extremes, is treble the difference between the first and the second. 17. This may be effected in different ways, one of which is the following. At one extremity A of the given line AB draw AC making any acute angle with AB and join BC; at any point D in BC draw DEF parallel to AC cutting AB in E and such that EF is equal to ED, draw FC cutting AB in G. Then AB is harmonically divided in E, G. 18. In the figure Euc. vi. 13. If E be the middle point of AC; then AE or EC is the arithmetic mean, and DB is the geometric mean, between AB and BC. If DE be joined and BF be drawn perpendicular on DE; then DF may be proved to be the harmonic mean between AB and BC. 19. In the fig. Euc. vi. 13. DB is the geometric mean between AB and BC, and if AC be bisected in E, AE or EC is the Arithmetic mean. The next is the same as-To find the segments of the hypotenuse of rihgt-angled triangle made by a perpendicular from the right angle, S having given the difference between half the hypotenuse and the perpendicular. 20. Let the line DF drawn from D the bisection of the base of the triangle ABC, meet AB in E, and CA produced in F. Also let AG drawn parallel to BC from the vertex A, meet DF in G. Then by means of the similar triangles; DF, FE, FG, may be shewn to be in harmonic progression. 21. If a triangle be constructed on AB so that the vertical angle is bisected by the line drawn to the point C. By Euc. vi. A, the point required may be determined. 22. Let DB, DE, DCA be the three straight lines, fig. Euc. III. 37; let the points of contact B, E be joined by the straight line BC cutting DA in G. Then BDE is an isosceles triangle, and DG is a line from the vertex to a point G in the base. And two values of the square on BD may be found, one from Theo. 37, p. 118: Euc. 111. 35; 11. 2; and another from Euc. 111. 36; II. 1. From these may be deduced, that the rectangle DC, GA, is equal to the rectangle AD, CG. Whence the, &c. 23. Let ABCD be a square and AC its diagonal. On AC take AE equal to the side BC or AB: join BE and at E draw EF perpendicular to AC and meeting BC in F. Then EC, the difference between the diagonal AC and the side AB of the square, is less than AB; and CE, EF, FB may be proved to be equal to one another: also CE, EF are the adjacent sides of a square whose diagonal is FC. On FC take FG equal to CE and join EG. Then, as in the first square, the difference CG between the diagonal FC and the side EC or EF, is less than the side EC. Hence EC, the difference between the diagonal and the side of the given square, is contained twice in the side BC with a remainder CG: and CG is the difference between the side CE and the diagonal CF of another square. By proceeding in a similar way, CG, the difference between the diagonal CF and the side CE, is contained twice in the side CE with a remainder: and the same relations may be shewn to exist between the difference of the diagonal and the side of every square of the series which is so constructed. Hence, therefore, as the difference of the side and diagonal of every square of the series is contained twice in the side with a remainder, it follows that there is no line which exactly measures the side and the diagonal of a square. 24. Let the given line AB be divided in C, D. On AD describe a semicircle, and on CB describe another semicircle intersecting the former in P; draw PE perpendicular to AB; then E is the point required. 25. Let AB be equal to a side of the given square. On AB describe a semicircle; at A draw AC perpendicular to AB and equal to a fourth proportional to AB and the two sides of the given rectangle. Draw CD parallel to AB meeting the circumference in D. Join AD, BD, which are the required lines. 26. Let the two given lines meet when produced in A. At A draw AD perpendicular to AB, and AE to AC, and such that AD is to AE in the given ratio. Through D, E, draw DF, EF, respectively parallel to AB, AC and meeting each other in F. Join AF and produce it, and the perpendiculars drawn from any point of this line on the two given lines will always be in the given ratio. 27. The angles made by the four lines at the point of their divergence, remain constant. See Note on Euc. vi. A, p. 295. 28. Let AB be the given line from which it is required to cut off a part BC such that BC shall be a mean proportional between the remainder AC and another given line. Produce AB to D, making BD equal to the other given line. On AD describe a semicircle, at B draw BE perpendicular to AD. Bisect BD in O, and with center O and radius OB describe a semicircle, join OE cutting the semicircle on BD in F, at F draw FC perpendicular to OE and meeting AB in C. C is the point of division, such that BC is a mean proportional between AC and BD. 29. Find two squares in the given ratio, and if BF be the given line (figure, Euc. vi. 4), draw BE at right angles to BF, and take BC, CE respectively equal to the sides of the squares which are in the given ratio. Join EF, and draw CA parallel to EF: then BF is divided in A as required. 30. Produce one side of the triangle through the vertex and make the part produced equal to the other side. Bisect this line, and with the vertex of the triangle as center and radius equal to half the sum of the sides, describe a circle cutting the base of the triangle. 31. If a circle be described about the given triangle, and another circle upon the radius drawn from the vertex of the triangle to the center of the circle, as a diameter, this circle will cut the base in two points, and give two solutions of the problem. Give the Analysis. 32. This Problem is analogous to the preceding. 33. Apply Euc. vi. 8, Cor.; 17. 34. Describe a circle about the triangle, and draw the diameter through the vertex A, draw a line touching the circle at A, and meeting the base BC produced in D. Then AD shall be a mean proportional between DC and DB. Euc. I. 36. 35. In BC produced take CE a third proportional to BC and AC; on CE describe a circle, the center being O; draw the tangent EF at E equal to AC; draw FO cutting the circle in T and T; and lastly draw tangents at T, T meeting BC in P and P'. These points fulfil the conditions of the problem. By combining the proportion in the construction with that from the simular triangles ABC, DBP, and Euc. III. 36, 37: it may be proved that CA, PDCP2. The demonstration is similar for P'D'. 36. This property may be immediately deduced from Euc. vI. 8, Cor. 37. Let ABC be the triangle, right-angled at C, and let AE on AB be equal to AC, also let the line bisecting the angle A, meet BC in D. Join DE. Then the triangles ACD, AED are equal, and the triangles ACB, DEB equiangular. 38. The segments cut off from the sides are to be measured from the right angle, and by similar triangles are proved to be equal; also by similar triangles, either of them is proved to be a mean proportional between the remaining segments of the two sides. 39. First prove AC: AD :: BC: 2. BD: then 2. AC: AD2:: BC: BD, whence 2. AC - AD: AD':: BC - BD: BD, and since 2. AC - AD = 2. AC - (AC2 + DC”) = AC2 — CD3, the property is immediately deduced. 40. The construction is suggested by Euc. 1. 47, and Euc. vi. 31. 41. See Note Euc. vi. A. p. 295. The bases of the triangles CBD, ACD, ABC, CDE may be shewn to be respectively equal to DB, 2.BD, 3.BD, 4.BD. 42. (1) Let ABC be the triangle which is to be bisected by a line drawn parallel to the base BC. Describe a semicircle on AB, from the center D draw DE perpendicular to AB meeting the circumference in E, join EA, and with center A and radius AE describe a circle cutting AB in F, the line drawn fron F parallel to BC, bisects the triangle. The |