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Bisect BC in E and join AE, DE, AF,
then the triangle ABE is half of the triangle ABC:
hence the triangle ABE is equal to the triangle DBF;

take away from these equals the triangle DBE,
therefore the remainder ADE is equal to the remainder DEF.

But ADE, DEF are equal triangles upon the same base DE, and on the same side of it,

they are therefore between the same parallels,

that is, AF is parallel to DE,

therefore the point F is determined.
Synthesis. Bisect the base BC in E, join DE,
from A, draw AF parallel to DE, and join DF.

Then because DE is parallel to AF,
therefore the triangle ADE is equal to the triangle DEF;

to each of these equals, add the triangle BDE, therefore the whole triangle ABE is equal to the whole DBF,

but ABE is half of the whole triangle ABC; therefore ABF is also half of the triangle ABC.

PROBLEM IV.

Given one angle, a side opposite to it, and the sum of the other two sides, construct the triangle.

Analysis. Suppose BAC the triangle required, having BC equal to the given side, BĀC equal to the given angle ppposite to BC, also BD equal to the sum of the other two sides.

D

A

Join DC.
Then since the two sides BA, AC are equal to BD,

by taking BA from these equals, the remainder AC is equal to the remainder A D. Hence the triangle ACD is isosceles, and therefore the angle ADC is equal to the angle ACD.

But the exterior angle BAC of the triangle ADC is equal to the two interior and opposite angles ACD and ADC:

Wherefore the angle BĂC is double the angle BDC, and BDC is the half of the angle BAC.

Hence the synthesis. At the point D in BD, make the angle BDC equal to half the given angle, and from B the other extremity of BD, draw BC equal to the given

side, and meeting DC in C, at C in CD make the angle DCA equal to the angle CDA, so that

CA may meet B D in the point A.
Then the triangle ABC shall have the required conditions.

PROBLEMS.

5. Given the base and one of the sides of an isosceles triangle, to describe the triangle.

6. Describe an isosceles triangle, each of the sides of which shall be double of the base.

7. In a given straight line, find a point equally distant from two given points; one in, and the other above or below, the given straight line.

8. AB, AC are straight lines cutting one another in A, D is a given point. Draw through D a straight line cutting off equal parts from AB and AC.

9. Draw through a given point, between two straight lines not parallel, a straight line, which shall be bisected in that point.

10. Divide a given right angle into three equal angles.

11. One of the acute angles of a right-angled triangle is three times as great as the other; trisect the smaller of these.

12. From a given point in a given straight line, it is required to erect a perpendicular by the help of straight lines only.

13. From a given point without a given straight line, to draw a line making an angle with the given line equal to a given rectilineal angle.

14. Through a given point draw a straight line which shall make equal angles with two straight lines given in position.

15. To determine that point in a straight line from which the straight lines drawn to two other given points shall be equal, provided the line joining the two given points is not perpendicular to the given line.

16. To place a straight line in a triangle (terminated by the two sides) which shall be equal to one straight line and parallel to another.

17. Determine the shortest path from one given point to another, subject to the condition, that it shall meet two given lines.

18. In the base of a triangle, find the point from which lines, drawn parallel to the sides of the triangle and limited by them, are equal.

19. From a given point in either of the equal sides of an isosceles triangle, to draw a straight line to the other side produced, which shall make with these sides a triangle equal to the given triangle. Prove that the line thus drawn will be greater than the base of the isosceles triangle.

20. From one of the obtuse angles of a rhomboid draw a straight line to the opposite side, which shall be bisected by the diagonal drawn through its acute angles.

21. Upon a straight line as a diagonal describe a parallelogram having an angle equal to a given angle.

22. On a given line to describe a parallelogram, having two of its opposite angles double of the other two, and all its sides equal. By means of this problem, trisect a right angle.

23. Describe a parallelogram equal to a given square, and having an angle equal to half a right angle.

24. To describe a rhombus which shall be equal to any given quadrilateral figure.

Describe a parallelogram equal in area and perimeter to a given triangle.

26. Describe on a given straight line a triangle which shall be equal to a given rectilineal figure, and have its vertical angle equal to a given angle.

27. Transform a given rectilinear figure into a triangle whose vertex shall be in a given angle of the figure and whose base be in one of the sides.

28. Straight lines are drawn from a fixed point to the several points of a straight line given in position, and on each base is described an equilateral triangle. Determine the locus of the vertices.

29. Upon a given base to describe an isosceles triangle, which shall be equal to a given triangle.

30. Given the base and one side of a triangle, to find the third side, so that the area may be the greatest possible.

31. Describe a right-angled triangle upon a given hypothenuse, so that the hypothenuse and one side shall be together double of the third side.

32. Having given two lines, which are not parallel, and a point between them; describe a triangle having two of its angles in the respective lines, and the third at the given point; and such that the sides shall be equally inclined to the lines which they meet.

33. Bisect a triangle by a straight line drawn parallel to one of its sides.

34. Bisect a triangle by a straight line drawn through a point within or without the triangle.

35. It is required to bisect any triangle by a line perpendicular to the base. 36. It is required to determine a point within a given triangle

, from which lines drawn to the several angles, will divide the triangle into three equal parts.

37. Bisect a parallelogram, (1) by a line drawn from a point in one of its sides : (2) by a line drawn from a given point within or without it: (3) by a line perpendicular to one of the sides.

38. To bisect a trapezium (1) by a line drawn from one of its angular points: (2) by a line drawn from a given point in one side.

39. Divide a triangle into three equal parts, (1) by lines drawn from a point in one of the sides: (2) by lines drawn from the angles to a point within the triangle: (3) by lines drawn from a given point within the triangle. In how many ways can the third case be done?

40. To trisect a parallelogram by lines drawn from a given point in one of its sides.

41. To divide a parallelogram into four equal portions by straight lines drawn from a given point in one of its sides.

42. To divide a square into four equal portions by three straight lines drawn from any point in one of its sides.

43. From a given isosceles triangle, cut off a trapezium which shall have the same base as the triangle, and shall have its remaining three sides equal to each other.

44. Divide an equilateral triangle into nine equal parts.

45. To find a point in the side or side produced of any parallelogram, such that the angle it makes with the line joining the point and one extremity of the opposite side, may be bisected by the line joining it with the other extremity.

46. Find a point in the diagonal of a square produced, from which if a straight line be drawn parallel to any side of the square, and meeting another side produced, it will form together with the produced diagonal and produced side, a triangle equal to the square.

47. A trapezium is such, that the perpendiculars let fall on a diagonal from the opposite angles are equal. Divide the trapezium into four equal triangles, by straight lines drawn to the angles from a point within it.

48. Given one side of a right-angled triangle, and the difference between the hypothenuse and the sum of the other two sides, to construct the triangle.

49.. In a right-angled triangle, given the sums of the base and the hypothenuse, and of the base and the perpendicular; to determine the triangle.

50. Given half the perimeter and the vertical angle of an isosceles triangle, it is required to find the sides.

51. Given the perimeter and the angles of a triangle, to construct it.

52. Given one of the angles at the base of a triangle, the base itself, and the sum of the two remaining sides, to construct the triangle.

53. Given the base, an angle adjacent to the base, and the difference of the sides of a triangle, to construct it.

54. Given one angle, a side opposite to it, and the difference of the other two sides; to construct the triangle.

55. Given the base, the perpendicular and the sum of the sides ; to construct the triangle.

56. Given the base, the altitude, and the difference of the two remaining sides; construct the triangle.

57. To find a point in the base of a triangle, such that if perpendiculars be drawn from it upon the sides, their sum shall be equal to a given line.

58. Determine the locus of the vertices of all the equal triangles, which can be described on the same base, and upon the same side of it.

59. To describe a square upon a given straight line as a diameter.

60. Shew how the squares upon the sides of a right-angled triangle may be dissected, so as exactly to cover the square of the hypothenuse. 61. Find a square equal to 3, 5, or any number of squares.

62. Construct a square whose area shall be 8 or n times that of a given square.

63. Construct all right-angled triangles whose sides shall be rational, upon a given line as their base.

64. Describe a square which shall be equal to the difference between two given squares.

65. In a right-angled triangle, it is required to find analytically, the base and perpendicular, their difference being 1, and the hypothenuse equal to 5.

66. Any two parallelograms having been described upon two sides of a given triangle, apply to the third side a parallelogram equal

67. Given a square of one inch, shew how a rhombus may structed, whose area shall be equal to it, and each of its sides a mile

their sum.

be con

long.

THEOREMS.

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2. In the fig. 1. 5. If FC and BG meet in H, then prove that AH bisects the angle BAC.

3. In the fig. 1. 5. If the angle FBG be equal to the angle ABC, and BG, CF intersect in 0; the angle BOF is equal to twice the angle BAC.

4. If a straight line drawn bisecting the vertical angle of a triangle, also bisects the base, the triangle is isosceles.

5. In the base BC of an isosceles triangle ABC take a point D, and in CA take CE equal to CD, let ED produced meet AB produced in F; then 3. AEF = 2 right angles + AFE.

6. The difference between any two sides of a triangle is less than the third side.

7. ABC is a triangle right-angled at B, and having the angle A double the angle C; shew that the side AC is less than double the side AB.

8. The difference of the angles at the base of any triangle, is double the angle contained by a line drawn from the vertex perpendicular to the base, and another bisecting the angle at the vertex.

9. If from the right angle of a right-angled triangle two straight lines be drawn, one perpendicular to the base, and the other bisecting it, they will contain an angle equal to the difference of the two acute angles of the triangle.

10. If one angle at the base of a triangle be double of "the other, the less side is equal to the sum or difference of the segments of the base made by the perpendicular from the vertex, according as the angle is greater or less than a right angle.

11. If one angle of a triangle be equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle.

12. If two exterior angles of a triangle be bisected, and from the point of intersection of the bisecting lines a line be drawn to the opposite angle of the triangle, it will bisect that angle.

13. In an obtuse-angled triangle if perpendiculars be drawn from the points bisecting the sides, prove that they all will pass through the same point.

14. In an obtuse-angled triangle, if perpendiculars be drawn from the angles to the opposite sides, produced if necessary, they will pass through the same point: required a proof.

15. Let ACB be a triangle; and let AD, CG, BE respectively, bisect the exterior angles HÅE, ECB, CBG, and meet BC, AB, AC produced in the points D, E, G. It is required to demonstrate that these three points are in the same straight line.

16. If two sides of a triangle be produced, the three straight lines which bisect the two exterior angles and the third interior angle shall all meet in the same point.

17. ABC is a triangle in which the angle ABC is twice the angle ACB; shew that if the point D in BC which divides it into segments, whose difference is equal to the side opposite to the angle ACD, be joined with the point Å, AD is perpendicular to BC.

18. In Prop. 35, Book 1, shew that the two parallelograms can be

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