25. Given L, L1, K; to construct a circle having a given radius, touching K, and having the distances from its centre to L and L1 in a given ratio.

26. Given P, L to L1, and L, cutting L and L1; through P to draw a line from which the given lines shall cut two segments having a given ratio.

27. In K given P and P1; to place in K a chord of given length so that the distances from P and P1 to the chord shall be as 3: 1.

To find a point X such that the distances from two given points shall have a given ratio, and

28. X shall be at the distance a from L.

29. X shall be equidistant from L and L.

30. The tangent from X to K shall have a given length. 31. The distances from X to two other given points shall also have a given ratio.

32. In a given triangle, to find a point the distances of which from the three vertices shall be as the numbers m:n: p.

33. To find a point from which the lengths AB, BC, CD taken in a straight line shall be seen under equal angles. (No. 143, and § 15, Ex. 12.)

34. To find a point from which three given circles will be seen under equal angles.

The distances of the required point from the centres of the circles are proportional to the respective radii of the circles. (Ex. 16, § 15.)

35. In a diameter of a circular billiard table are placed two balls; to find by construction in what direction one of the balls must be struck centrally in order that it may hit the other after first striking the side of the table.

36. To find the point which is equally illuminated by three lights situated in the same plane. (Ex. 10, § 15.)

37. To inscribe a square in a semicircle.

Suppose the problem solved, and ABCD (Fig. 78) the square re

38. To inscribe a square in a given triangle.

Suppose the problem solved, and DEFG (Fig. 79) the inscribed square. Draw CM to AB, and let AF produced meet CM in M; then GF: CM= AF: AM. Draw MNL to AB, and CH 1 to AB; then FE: MN = AF: AM. Now GF FE; .. CM = MN - CH. Therefore, construct a square upon CH as one side, and the line AM will determine the point F.

=

39. To inscribe in a triangle a rectangle similar to a given rectangle.

The solution is like that of Ex. 38; upon CH (Fig. 79) construct a rectangle similar to the given rectangle.

40. To inscribe in a semicircle a rectangle similar to a given rectangle.

41. To inscribe in a circle a triangle similar to a given triangle.

42. To circumscribe about a circle a triangle similar to a given triangle.

In the triangle ABC to draw a parallel to AB, meeting AC in X and BC in Y, so that,

43. AB: XY: =m: N. 44. AC: XY:

= XY: CX.

From the similar ▲ ABC, XYC we have AB: AC = XY: CX;

through P to draw a chord XY

Through Cdraw a line I to AB; produce A Yand BX to meet it. 49. Given P within K; so that PX: PY=m: n. 2 inches, between what limits must the ratio m:n lie in order that the problem may be possible?

If 4 inches, PO

=

Draw AY to OX and meeting OP in A; then OP: OA = m : n, also r: AY =m: n.

50. Given Pin the arc AB of K; to draw a chord PX which shall be divided by the chord AB in the ratio m: n. Draw XC to AB and meeting PO.

51. Given K, a chord AB, P in AB; through P to draw a chord XY so that AX: BY =m: N.

52. Given Kand a chord AB; to find a point Xin K so that AX: BX = m : n.

53. In K to draw a chord which shall be divided by two given radii into three equal parts.

54. Given P in K; from P to draw two chords PX, PY so that PX: PY= m: n, and XY shall be a diameter. Through P draw a line 1 to PO; two similar ▲ are formed.

55. In a given arc AB to find a point X such that AXX BX = k2.

Given P without K; through P to draw a secant PXY,

Draw a tangent PA, also BX || to AY; then PA is divided by B in extreme and mean ratio; also from the similar & PAX, PBX we have PX2 PA × PB.

=

61. Given K and two radii OA, OB; to draw a tangent CXD, bisecting Kin X and meeting OA, OB produced in C, D, so that CX: XD

=m: n.

Draw AE to CD, meeting OX in E and OD in F, and draw CG || to OD meeting OA in G; then AG: GO =m: N.

62. Given K and K, intersecting in P; through P to draw a line so that the chords intercepted by the circles shall be as m: n.

A line through P perpendicular to the required line will divide the line of centres in the ratio mn.

63. In L to find a point X from which the tangents drawn to K and K, shall be equal.

64. To find a point X from which the tangents drawn to three given circles shall be equal.

65. To construct a circle having a given radius, and cutting two given circles at right angles.

66. To construct a circle which shall pass through Pand cut K at right angles in P1.

67. To construct a circle which shall pass through Pand cut K and K, at right angles.

68. To construct a circle which shall cut three given circles at right angles.

§ 17. THE METHOD OF SIMILAR FIGURES.

1. This method of constructing figures is based upon the fact that the shape of a figure is determined by certain angles or the ratios of certain lines. First construct a figure similar to the figure required, and then construct the required figure by means of the general theorem that in similar figures homologous lines are proportional.

2. The shape of a triangle'is determined by: (i.) two angles;

(ii.) the ratio of two sides and the included angle; (iii.) the ratios of the three sides;

(iv.) the ratios of the three altitudes;

(v.) the ratio of an altitude to the corresponding base, and the angle at the vertex.

3. This method is also applicable to cases in which the shape, not of the required figure, but of an auxiliary triangle, is determined by the given parts; as, for example, by: (i.) one angle;

(ii) the ratio of an altitude to one of the adjacent sides; (iii) the ratio of an altitude to the corresponding median; (iv.) the ratio of the radius of the circumscribed circle to one side.

4. The notation used in the following exercises is the same as in Chapter II.

5. To construct a triangle, given a, B, h+m.

Analysis. Any ▲ DEC(Fig. 80), in which ▲ CDE= a, ▲ CED= ß, will be similar to the triangle required. If in this triangle we draw. the altitude CF and the median CG, then will each side of the required triangle be a fourth proportional to three known lengths, CF+CG, h+m, and the homologous side of the ▲ DEC. The problem is then reduced to the construction of a triangle from two angles and one side.