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Let this circle be described, and let O be its center; join 40, OD: then OD is at right angles to AE, (Euc. III. 3.)

and ADO is a right-angled triangle.

Hence the semicircle described on 40 as a diameter will pass through the point D, (Euc. III. 31.)

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Therefore the point D is determined to be that point in the base, where the circumference of the semicircle described upon the radius of the circumscribing circle of the triangle, intersects the base.

If the circle be completed, it will intersect the base in another point G, so that AG is also a mean proportional between BG and G.C.

Synthesis. Describe a circle about the given triangle ABC, and let O be the center: join 40, and on 40 as a diameter describe a semicircle ADO intersecting the base BC in D, join AD, then AD is a mean proportional between AD and DB.

Produce AD to meet the circumference in E, and join OD. Then because OD is at right angles to AE, and is drawn from the center 0,

therefore AD is equal to DE. (Euc. III. 3.)

And because AE, BC intersect each other within the circle; the rectangle BD, DC is equal to the rectangle AD, DE, (Euc. III. 35.) but DE is equal to AD,

therefore the rectangle BD, DC is equal to the square on AD: and AD is a mean proportional between BD and DČ. (Euc. VI. 17.) If AG be joined, then AG can be shewn to be a mean proportional between BG and GC; which is a second solution of the Problem.

Cor. 1. If the given triangle ABC become right-angled at A, the center of the circle will bisect the base BC, and the lines AD, AG will coincide at the second point D, where the circle cuts the base BC, and be at right angles to BC, and AD will be a mean proportional between BD and DC.

COR. 2. If the triangle ABC be acute-angled at A, the problem is impossible; as it is obvious that the circle which circumscribes the triangle ABC, will have its center ✪ within the triangle, and the semicircle described upon the radius 40 as a diameter cannot intersect the base BC of the triangle.

If, however, a tangent AD be drawn to touch the circumscribing circle at the point 4, and CB be produced to meet the tangent AD in the point D the rectangle contained by DC, DB is equal to the square on DA: (Euc. III. 36.) and DA is a mean proportional between DC and DB (Eue. vI. 17.) Hence a straight line AD has been drawn from 4, an angle of an acute-angled triangle to meet the base BC produced in D, so that AD is a mean proportional between CD and DB.

I.

7. If ACB, ADB be two triangles upon the same base AB, and between the same parallels, and if through the point in which two of the sides (or two of the sides produced) intersect two straight lines, be drawn parallel to the other two sides so as to meet the base AB (or AB produced) in points E and F. Prove that AE= BF.

8. In the base AC of a triangle ABC take any point D; bisect AD, DC, AB, BC, in E, F, G, H respectively: shew that EG is equal to HF.

9. If, in similar triangles, from any two equal angles to the opposite sides, two straight lines be drawn making equal angles with the homologous sides, these lines will have the same ratio as the sides on which they fall, and will also divide those sides proportionally.

10. BD, CD are perpendicular to the sides AB, AC of a triangle ABC, and CE is drawn perpendicular to AD, meeting AB in E: shew that the triangles ABC, ACE are similar.

11. In any triangle, if a perpendicular be let fall upon the base from the vertical angle, the base will be to the sum of the sides, as the difference of the sides to the difference or sum of the segments of the base made by the perpendicular, according as it falls within or without the triangle.

12. If two triangles on the same base have their vertices joined by a straight line which meets the base or the base produced; the parts of this line between the vertices of the triangles and the base, are in the same ratio to each other as the areas of the triangles.

13. In the triangle ABC there are drawn AD bisecting BC, and EF parallel to BC and cutting AB in E and AC in F. Shew that BF and CE will intersect in AD.

14. If the side BC of a triangle ABC be bisected in D, and the angles ADB, ADC be bisected by the straight lines DE, DF, meeting AB, AC in E, F respectively, shew that EF is parallel to BC.

15. APB, CQD are two parallel right lines, and AP is to PB as DQ is to QC, prove that the right lines PQ, AD, BC meet in a point. 16. CAB, CEB are two triangles having a common angle CBA, and the sides opposite to it CA, CE, equal; if BAE be produced to D, and ED be taken a third proportional to BA, AC, then shall the triangle BDC be similar to the triangle BAC.

17. From a point E in the common base of two triangles ACB, ADB, straight lines are drawn parallel to AC, AD meeting BC, BD in Fand G, shew that the lines joining F, G and C, D will be parallel. 18. AB is a given straight line, and D a given point on' it; it is required to find a point P, in AB produced, such that AP is to PB as AD is to DB.

19. ABb, AcС are two given straight lines, cut by two others BC, bc, so that the two triangles ABC, Abc may be equal; the lines BC, bc, divide each other proportionally.

20. If AB, AC be two given straight lines, and AD be taken on AB, and AE on AC, such that AB = m. AD, and AC=m. AE; join BE, CD to meet in F; shew that DE= (m + 1). DF.

21. ABC is a triangle; Ab, Ac are taken on AB, AC respectively, such that AB-n. Ab, and AC=n. Ac; Cb, Be meet in D, and AD produced meets BC in E: shew that 2.AE= (n+1). AD, and E is the middle point of BC.

22. AB, AC are two straight lines, B and C given points in the

same; BD is drawn perpendicular to AC and DE perpendicular to AB; in like manner CF is draw perpendicular to AB, and FG to AC, shew that EG is parallel to BC.

23. Lines drawn from the extremities of the base of a triangle intersecting in the line joining the vertex with the point of bisection of the base, cut the sides proportionally: and conversely.

24. ABC is any triangle, D any point in AB produced; E the point in BC, such that CE: EB:: AD: BD. Prove that DE produced will bisectTM A C.

25. In a given triangle draw a straight line parallel to one of the sides so that it may be a mean proportional to the segments of the base. 26. Through E, F, two points on the side AB of a triangle ABC, so taken that AE is equal to BF, two straight lines EG, BH are drawn respectively parallel to the other sides and meeting them in G and H; if GH be joined, it will be parallel to AB.

27. If the three sides of a triangle be bisected, the lines which join the points of bisection will divide the triangle into four equal triangles, each of them similar to the whole triangle.

28. If the side BC of a triangle ABC be bisected in D, and the angles ADB, ADC be bisected by the straight lines DE, DF, meeting AB, AC in E, Frespectively, shew that EF is parallel to BC.

29. A straight line drawn through the middle point of one side of a triangle divides the two other sides, the one internally and the other externally in the same ratio.

30. If two of the interior angles of a triangle ABC be bisected by the lines COE, BOD intersecting in 0, and meeting the opposite sides in the points E and D, prove that

OD: OB:: AD: AB, and OC: OE:: AC: AE.

31. If the vertical angle CAB of a triangle ABC be bisected by AD, to which the perpendiculars CE, BF, are drawn from the remaining angles; bisect the base BC in G, join GE, GF, and prove these lines equal to each other.

32. If a side BC of a triangle ABC be bisected by a straight line which meets the sides AB, AČ (produced if necessary) in D and E respectively; the line AE will be to EC, as AD to DB.

33. The base AB of an isosceles triangle ABC is produced both ways to A' and B', so that AA'. BB′ = AČ2; shew that the triangles. AAC, BBC are similar to one another.

34. To find a point P in the base BC of a triangle produced, so that PD being drawn parallel to AC, and meeting AC produced to D, AC: CP:: CP:PD.

35. If from the extremities of the base of a triangle, two straight lines be drawn, each of which is parallel to one of the sides, and equal to the other, the straight lines joining their other extremities with the other extremities of the base, will cut off equal segments from the sides, and each of these will be a mean proportional between the other two segments.

36. A straight line CD is drawn bisecting the vertical angle C of a triangle ACB, and cutting the base AB in D; also on AB produced a point E is taken equidistant from C and D: prove that AE.BE=DE2.

37. ABC is an equilateral triangle; E any point in AC; in BC produced take CD= CA, CF = CE, AF, DE intersect in H; then HC:EC:: AC: AC + EC.

38. ABC is an isosceles triangle; draw CE perpendicular to the base AB; draw ADF intersecting CE in D, and CB in F; then DE: CE:: CA - EF : CA + CĘ.

39. From an angle of a triangle a line is drawn to the middle point of the opposite side. And through the point of bisection of this line another is drawn from either angle to the side subtending it. Prove that the latter line divides this side into segments which are as 2 to 1.

40. Determine the point in the produced side of a triangle, from which a straight line being drawn to a given point in the base, shall be intersected by the other side of the triangle in a given ratio.

41. If from two points P, Q, four perpendiculars be dropped upon the straight lines AB, AC, such that the perpendiculars are proportionals; shew that P and Q lie in the same straight line through A. 42. If one side of a triangle be produced, and the other shortened by equal quantities, the line joining the points of section will be divided by the base in the inverse ratio of the sides.

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43. If the triangle ABC has the angle at C a right angle, and from C a perpendicular be dropped on the opposite side intersecting it in D, then AD: DB:: AC2: CB2.

44. In any right-angled triangle, one side is to the other, as the excess of the hypotenuse above the second, to the line cut off from the first between the right angle and the line bisecting the opposite angle.

45. If on the two sides of a right-angled triangle squares be described, the lines joining the acute angles of the triangle and the opposite angles of the squares, will cut off equal segments from the sides; and each of these equal segments will be a mean proportional between the remaining segments.

46. ABC is a right-angled triangle, having a right angle at C; find a point P in the hypotenuse, so situated, that PA may be half of the perpendicular, dropped from P upon BC the base.

47. Determine that point in the base produced of a right-angled triangle from which the line drawn to the angle opposite the base, shall have the same ratio to the base produced, which the perpendicular

has to the base itself.

48. If the perpendicular in a right-angled triangle divide the hypotenuse in extreme and mean ratio, the less side is equal to the alternate segment.

49. From B the right angle of a right-angled triangle ABC, Bp is let fall perpendicular to AC, from p, pq is let fall perpendicular to BA, &c.: prove that

Bp+pq+ &c.: AB:: AB+ AC: BC. 50. If the triangle ACB has a right angle C, and AD be drawn bisecting the angle 4, and meeting CB in D, prove that

51.

AC2: AD :: BC:2.BD.

In any right-angled triangle ABC, (whose hypotenuse is AB) bisect the angle A by AD meeting CB in D, and prove that

2A C2: A C12 - CD2:: BC: CD.

52. ABC is a right-angled triangle, CD a perpendicular from

the right angle upon AB; show that if AC is double of BC, BD is one-fifth of AB.

53. If through C the extremity of the hypotenuse BC of a right-angled triangle ABC there be drawn two straight lines bisecting the internal and external angles at C, and meeting BA produced in D and E respectively, AC will be a mean proportional between AD and AE.

54. If F be a point in the side CB, and CD, FE be perpendiculars on the hypotenuse AB, then AD. AE+ CD . EF=AC2.

III.

55. Triangles and parallelograms of unequal altitudes are to each other in the ratio compounded of the ratios of their bases and altitudes. 56. If triangles AEF, ABC have a common angle A, triangle ABC: triangle AEF:: AB.AC: AE. AF.

57. Construct an isosceles triangle equal to a given scalene triangle and having an equal vertical angle with it.

58. On two given straight lines similar triangles are described. Required to find a third, on which, if a triangle similar to them be described, its area shall equal the difference of their areas.

59. In the triangle ABC, AC=2. BC. If CD, CE respectively bisect the angle C, and the exterior angle formed by producing A0; prove that the triangles CBD, ACD, ABC, CDE, have their areas as 1, 2, 3, 4.

60. If two sides of a triangle be bisected and the points joined with the opposite angles, the joining lines shall divide each other proportionally. And the triangle formed by the joining lines and the remaining side shall be equal to a third of the original triangle.

61. A square is inscribed below the base of an isosceles triangle, prove that if the vertex be joined to the corners of the square, the middle segment of the base will be to the outer one in double the ratio of the perpendicular on the base to the base.

6.2. Dis the middle point of the base BC of an isosceles triangle, CF perpendicular to AB, DE perpendicular to CF, EG parallel to the base meets AD in G; prove that EG is to GA in the triplicate ratio of BD to DA.

63. If a straight line be drawn through the points of bisection of any two sides of a triangle, it will divide the triangle into two parts which are to each other as 1 to 3.

64. If perpendiculars be drawn from the extremities of the base to a triangle on a straight line which bisects the angle opposite to the base, the area of the triangle is equal to the rectangle contained by either of the perpendiculars, and the segment of the bisecting line between the angle and the other perpendicular.

65. If the angles at the base of an isosceles triangle (which are double of the vertical angle) be bisected by lines meeting the opposite sides, and the points of intersection with the sides be joined, then the joining line will divide the triangle into two parts, which have the same constant ratio to one another, as one of the sides of the triangle bears to the base.

66. The square on the line bisecting the vertical angle of any triangle is a mean proportional between the differences of the squares on each side containing that angle, and the square on the adjacent segment of the base.

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