164. Let AB be the given straight line. Bisect AB in C and on AB as a diameter describe a circle; and at any point D in the circumference, draw a tangent DE equal to a side of the given square; join DC, EC, and with center C and radius CE describe a circle cutting AB produced in F. From F draw FG to touch the circle whose center is C in the point G. 165. Let AD, DF be two lines at right angles to each other, O the centre of the circle BFQ; A any point in AD from which tangents AB, AC are drawn; then the chord BC shall always cut FD in the same point P, wherever the point A is taken in AD. Join AP; then BAC is an isosceles triangle, and FD. DE + AD2 = AB2 = BP. PC + AP2 = BP. PC + AD2 + DP2, wherefore BP. PC FD. DE - DP. = The point P, therefore, is independent of the position of the point A ; and is consequently the same for all positions of A in the line AD. 166. The point E will be found to be that point in BC, from which two tangents to the circles described on AB and CD as diameters, are equal, Euc. III. 36. 167. If AQ, A'P' be produced to meet, these lines with AA' form a right-angled triangle, then Euc. 1. 47. GEOMETRICAL EXERCISES ON BOOK IV. HINTS, &c. 1. LET AB be the given line. Draw through C the center of the given circle the diameter DCE. Bisect AB in F and join FC. Through A, B draw AG, BH parallel to FC and meeting the diameter in G, H: at G, H draw GK, HL perpendicular to DE and meeting the circumference in the points K, L ; join KL; then KL is equa' and parallel to AB. 2. Trisect the circumference and join the center with the points of trisection. 3. See Euc. IV. 4, 5. 4. Let a line be drawn from the third angle to the point of intersection of the two lines; and the three distances of this point from the angles may be shewn to be equal. 5. Let the line AD drawn from the vertex A of the equilateral triangle, cut the base BC, and meet the circumference of the circle in D. Let DB, DC be joined: AD is equal to DB and DC. If on DA, DE be taken equal to DB, and BE be joined; BDE may be proved to be an equilateral triangle, also the triangle ABE may be proved equal to the triangle CBD. The other case is when the line does not cut the base. 6. Let a circle be described upon the base of the equilateral triangle, and let an equilateral triangle be inscribed in the circle. Draw a diameter from one of the vertices of the inscribed triangle, and join the other extremity of the diameter with one of the other extremities of the sides of the inscribed triangle. The side of the inscribed triangle may then be proved to be equal to the perpendicular in the other triangle. 7. The line joining the points of bisection, is parallel to the base of the triangle and therefore cuts off an equilateral triangle from the given triangle. By Euc. III. 21; 1. 6, the truth of the theorem may be shewn. 8. Let a diameter be drawn from any angle of an equilateral tri angle inscribed in a circle to meet the circumference. It may be proved that the radius is bisected by the opposite side of the triangle. 9. Let ABC be an equilateral triangle inscribed in a circle, and let ABC be an isosceles triangle inscribed in the same circle, having the same vertex A. Draw the diameter AD intersecting BC in E, and B'C' in E, and let B'C' fall below BC. Then AB, BE, and AB', B'E', are respectively the semi-perimeters of the triangles. Draw B'F perpendicular to BC, and cut off AH equal to AB, and join BH. If BF can be proved to be greater than B'H, the perimeter of ABC is greater than the perimeter of AB'C'. Next let B'C' fall above BC. 10. The angles contained in the two segments of the circle, may be shewn to be equal, then by joining the extremities of the arcs, the two remaining sides may be shewn to be parallel. 11. It may be shewn that four equal and equilateral triangles will form an equilateral triangle of the same perimeter as the hexagon, which is formed by six equal and equilateral triangles. 12. Let the figure be constructed. By drawing the diagonals of the hexagon, the proof is obvious. 13. By Euc. 1. 47, the perpendicular distance from the center of the circle upon the side of the inscribed hexagon may be found. 14. The alternate sides of the hexagon will fall upon the sides of the triangle, and each side will be found to be equal to one-third of the side of the equilateral triangle. 15. A regular duodecagon may be inscribed in a circle by means of the equilateral triangle and square, or by means of the hexagon. The area of the duodecagon is three times the square on the radius of the circle, which is the square on the side of an equilateral triangle inscribed in the same circle. Theorem 1, p. 196. 16. In general, three straight lines when produced will meet and form a triangle, except when all three are parallel or two parallel are intersected by the third. This Problem includes Euc. Iv. 5, and all the cases which arise from producing the sides of the triangle. The circles described touching a side of a triangle and the other two sides produced, are called the escribed circles. 17. This is manifest from Euc. r. 21. 18. The point required is the center of the circle which circumscribes the triangle. See the notes on Euc. 11. 20, p. 155. 19. If the perpendiculars meet the three sides of the triangle, the point is within the triangle, Euc. IV. 4. If the perpendiculars meet the base and the two sides produced, the point is the center of the escribed circle. 20. This is manifest from Euc. II. 11, 18. 21. The base BC is intersected by the perpendicular AD, and the side AC is intersected by the perpendicular BE. From Theorem 1. p. 160; the arc AF is proved equal to AE, or the arc FE is bisected in A. In the same manner the arcs FD, DE, may be shewn to be bisected in BC. 22. Let ABC be a triangle, and let D, E be the points where the inscribed circle touches the sides AB, AC. Draw BE, CD intersecting each other in O. Join AO, and produce it to meet BC in F. Then Fis the point where the inscribed circle touches the third side BC. If F be not the point of contact, let some other point G be the point of contact, Through D draw DH parallel to AC, and DK parallel to BC. By the similar triangles, CG may be proved equal to CF, or G the point of contact coincides with F, the point where the line drawn from À through ✪ meets BC. 23. In the figure, Euc. Iv. 5. Let AF bisect the angle at A, and be produced to meet the circumference in G. Join GB, GC and find the center II of the circle inscribed in the triangle ABC. The lines GH, GB, GC are equal to one another. 24. Let ABC be any triangle inscribed in a circle, and let the perpendiculars AD, BE, CF intersect in G. Produce AD to meet the circumference in H, and join BH, CH. Then the triangle BHC may be shewn to be equal in all respects to the triangle BGC, and the circle which circumscribes one of the triangles will also circumscribe the other. Similarly may be shewn by producing BE and CF, &c. 25. First. Prove that the perpendiculars Aa, Bb, Cc pass through the same point O, as Theo. 112, p. 171. Secondly. That the triangles Acb, Bea, Cab are equiangular to ABC. Euc. 11. 21. Thirdly. That the angles of the triangle abc are bisected by the perpendiculars; and lastly, by means of Prob. 4, p. 71, that ab+be+ca is a minimum. 26. The equilateral triangle can be proved to be the least triangle which can be circumscribed about a circle. 27. Through C draw CH parallel to AB and join AH. Then HAC the difference of the angles at the base is equal to the angle HFC. Euc. 111. 21, and HFC is bisected by FG. 23. Let F, G, (figure, Euc.iv. 5,) be the centers of the circumscribed and inscribed circles; join GF, GA, then the angle GAF which is equal to the difference of the angles GAD, FAD, may be shewn to be equal to half the difference of the angles ABC and ACB. 29. This Theorem may be stated more generally, as follows: Let AB be the base of a triangle, AEB the locus of the vertex; D the bisection of the remaining arc ADB of the circumscribing circle; then the locus of the center of the inscribed circle is another circle whose center is D and radius DB. For join CD: then P the center of the inscribed circle is in CD. Join AP, PB; then these lines bisect the angles CAB, CBA, and DB, DP, DA may be proved to be equal to one another. 30. Let ABC be a triangle, having C a right angle, and upon AC, BC, let semicircles be described: bisect the hypotenuse in D, and let fall DE, DF perpendiculars on AC, BC respectively, and produce them to meet the circumferences of the semicircles in P, Q; then DP may be proved to be equal to DQ. 31. Let the angle BAC be a right angle, fig. Euc. Iv. 4. Join AD. Then Euc. III. 17, note p. 155. 32. Suppose the triangle constructed, then it may be shewn that the difference between the hypotenuse and the sum of the two sides is equal to the diameter of the inscribed circle. 33. Let P, Q be the middle points of the arcs AB, AC, and let PQ be joined, cutting AB, AC in DE; then AD is equal to AE. Find the center O and join OP, QO. 34. With the given radius of the circumscribed circle, describe a circle. Draw BC cutting off the segment BAC containing an angle equal to the given vertical angle. Bisect BC in D, and draw the diameter EDF: join FB, and with center F and radius FB describe a circle: this will be the locus of the centers of the inscribed circle (see Theorem 33, supra.) On DE take DG equal to the given radius of the inscribed circle, and through G draw GH parallel to BC, and meeting the locus of the centers in H. H is the center of the inscribed circle. 35. This may readily be effected in almost a similar way to the pre-ceding Problem. 36. With the given radius describe a circle, then by Euc. III. 34. 37. Let ABC be a triangle on the given base BC ana naving its vertical angle A equal to the given angle. Then since the angle at A is constant, A is a point in the arc of a segment of a circle described on BC. Let D be the center of the circle inscribed in the triangle ABC. Join DA, DB, DC: then the angles at B, C, A, are bisected. Euc. IV. 4. Also since the angles of each of the triangles ABC, DBC are equal to two right angles, it follows that the angle BDC is equal to the angle A and half the sum of the angles B and C. But the sum of the angles B and C can be found, because A is given. Hence the angle BDC is known, and therefore D is the locus of the vertex of a triangle described on the base BC and having its vertical angle at D double of the angle at A. 38. Suppose the parallelogram to be rectangular and inscribed in the given triangle and to be equal in area to half the triangle: it may be shewn that the parallelogram is equal to half the altitude of the triangle, and that there is a restriction to the magnitude of the angle which two adjacent sides of the parallelogram make with one another. 39. Let ABC be the given triangle, and A'B'C' the other triangle, to the sides of which the inscribed triangle is required to be parallel. Through any point a in AB draw ab parallel to A'B' one side of the given triangle and through a, b draw ac, be respectively parallel to AC, BC. Join Ac and produce it to meet BC in D; through D draw DE, DF, parallel to ca, cb, respectively, and join EF. Then DEF is the triangle required. 40. This point will be found to be the intersection of the diagonale of the given parallelogram. 41. The difference of the two squares is obviously the sum of the four triangles at the corners of the exterior square. 42. (1) Let ABCD be the given square: join AC, at A in AC, make the angles CAE, CAF, each equal to one-third of a right angle, and join EF. (2) Bisect AB any side in P, and draw PQ parallel to AD or BC, then at P make the angles as in the former case. 43. Each of the interior angles of a regular octagon may be shewn to be equal to three-fourths of two right angles, and the exterior angles made by producing the sides, are each equal to one fourth of two right angles, or one-half of a right angle. 44. Let the diagonals of the rhombus be drawn; the center of the inscribed circle may be shewn to be the point of their intersection. 45. Let ABCD be the required square. Join O, O' the centers of the circles and draw the diagonal AEC cutting OO' in E. Then E is the middle point of OO' and the angle AEO is half a right angle. 46. Let the squares be inscribed in, and circumscribed about a circle, and let the diameters be drawn, the relation of the two squares is manifest. 47. Let one of the diagonals of the square be drawn, then the isosceles right-angled triangle which is half the square, may be proved to be greater than any other right-angled triangle upon the same hypotenuse. 48. Take half of the side of the square inscribed in the given circle, this will be equal to a side of the required octagon. At the extremities on the same side of this line make two angles each equal to three-fourths of two right angles, bisect these angles by two straight lines, the point at which they meet will be the center of the circle which circumscribes the octagon, and either of the bisecting lines is the radius of the circle. 49. First shew the possibility of a circle circumscribing such a figure, and then determine the center of the circle. 50. By constructing the figures and drawing lines from the center of the circle to the angles of the octagon, the areas of the eight triangles may be easily shewn to be equal to eight times the rectangle contained by the radius of the circle, and half the side of the inscribed square. 51. Let AB, AC, AD, be the sides of a square, a regular hexagon and an octagon respectively inscribed in the circle whose center is O. Produce AC to E making AE equal to AB, from E draw EF touching the circle in F, and prove EF to be equal to AD. 52. Let the circle required touch the given circle in P, and the given line in Q. Let C be the center of the given circle and C' that of the required circle. Join CC, C'Q, QP; and let QP produced meet the given circle in R, join RC and produce it to meet the given line in V. Then RCV is perpendicular to VQ. Hence the construction. 53. Let A, B be the centers of the given circles and CD the given straight line. On the side of CD opposite to that on which the circles are situated, draw a line EF parallel to CD at a distance equal to the radius of the smaller circle. From A the center of the larger circle describe a concentric circle GH with radius equal to the difference of the radii of the two circles. Then the center of the circle touching the circle GH, the line EF, and passing through the center of the smaller circle B, may be shewn to be the center of the circle which touches the circles whose centers are A, B, and the line CD. 54. Let AB, CD be the two lines given in position and E the center of the given circle. Draw two lines FG, HI parallel to AB, CD respectively and external to them. Describe a circle passing through E and touching FG, HI. Join the centers E, O, and with center O and radius equal to the difference of the radii of these circles describe a circle; this will be the circle required. 55. Let the circle ACF having the center G, be the required circle touching the given circle whose center is B, in the point A, and cutting the other given circle in the point C. Join BG, and through A draw a line perpendicular to BG; then this line is a common tangent to the circles whose centers are B, G. Join AC, GC. Hence the construction. 56. Let C be the given point in the given straight line AB, and D the center of the given circle. Through C draw a line CE perpendicular to AB; on the other side of AB, take CE equal to the radius of the given circle. Draw ED, and at D make the angle EDF equal to the angle DEC, and produce EC to meet DF. This gives the construction for one case, when the given line does not cut or touch the other circle. 57. This is a particular case of the general problem; To describe a eircle passing through a given point and touching two straight lines given in position. Let A be the given point between the two given lines which when produced meet in the point B. Bisect the angle at B by BD and through A draw AD perpendicular to BD and produce it to meet the two given lines in C, E. Take DF equal to DA, and on CB take CG such that the rectangle contained by CF, CA is equal to the square on CG. The circle described through the points F, A, G, will be the circle required. Deduce the particular case when the given lines are at right angles to one another, and the given point in the line which bisects the angle at B. If the lines are parallel, when is the solution possible? 58. Let A, B, be the centers of the given circles, which touch externally in E; and let C be the given point in that whose center is B. Make CD equal to AE and draw AD; make the angle DAG equal to the angle ADG: then G is the center of the circle required, and GC its radius. |