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Also, since from the point F without the circle, are drawn two straight lines, one of which FE touches the circle, and the other FDC cuts it; the rectangle contained by FC, FD, is equal to the square on FE. (III. 36.)

Synthesis. Join C, D, and produce CD to meet AB in F, take the point E in FB, such that the square on FE, shall be equal to the rectangle FD, FC.

Bisect CD in H, and draw HK perpendicular to CD;
then HK passes through the center. (III. 1, Cor. 1.)
At E draw EG perpendicular to FB,

then EG passes through the center, (III. 19.) consequently L, the point of intersection of these two lines, is the center of the circle.

It is also manifest, that another circle may be described passing through C, D, and touching the line AB on the other side of the point F; and this circle will be equal to, greater than, or less than the other circle, according as the angle CFB is equal to, greater than, or less than the angle CFA.

PROPOSITION III.

PROBLEM.

Inscribe a circle in a given sector of a circle.

Analysis. Let CAB be the given sector, and let the required circle whose center is O, touch the radii in P, Q, and the arc of the sector in D.

[blocks in formation]

E

D

F

Join OP, OQ, these lines are equal to one another.
Join also CO.

Then in the triangles CPO, CQO, the two sides PC, CO, are equal to QC, CO, and the base OP is equal to the base OQ;

therefore the angle PCO is equal to the angle QCO;
and the angle ACB is bisected by CO:

also CO produced will bisect the arc AB in D. (III. 26.)
If a tangent EDF be drawn to touch the arc AB in D;
and CA, CB be produced to meet it in E, F:

the inscription of the circle in the sector is reduced to the inscrip-
tion of a circle in a triangle. (IV. 4.)

PROPOSITION IV.

ABCD is a rectangular parallelogram.

Required to draw EG, FG

parallel to AD, DC, so that the rectangle EF may be equal to the figure EMD, and EB equal to FD.

Analysis. Let EG, FG be drawn, as required, bisecting the rectangle ABCD.

PROBLEM.

Draw the diagonal BD cutting EG in H and FG in K.
Then BD also bisects the rectangle ABCD;

and therefore the area of the triangle KGH is equal to that of the two triangles EHB, FKD.

A

F

D

K

E B

[blocks in formation]

N C

Draw GL perpendicular to BD, and join GB,
also produce FG to M, and EG to N.

If the triangle LGH be supposed to be equal to the triangle EHB, by adding HGB to each,

the triangles LGB, GEB are equal, and they are upon the same base GB, and on the same side of it;

therefore they are between the same parallels,

that is, if L, E were joined, LE would be parallel to GB;

and if a semicircle were described on GB as a diameter, it would pass through the points E, L; for the angles at E, L are right angles:

also LE would be a chord parallel to the diameter GB; therefore the arcs intercepted between the parallels LE, GB are equal,

and consequently the chords EB, LG are also equal;
but EB is equal to GM, and GM to GN;
wherefore LG, GM, GN, are equal to one another;

hence G is the center of the circle inscribed in the triangle BDC. Synthesis. Draw the diagonal BD.

Find G the center of the circle inscribed in the triangle BDC; through G draw EGN parallel to BC, and FKM parallel to AB. Then EG and FG bisect the rectangle ABCD.

Draw GL perpendicular to the diagonal BD.

In the triangles GÌÍ, EHB, the angles GLH, HEB are equal, each being a right angle, and the vertical angles LHG, EHB, also the side LG is equal to the side EB;

therefore the triangle LHG is equal to the triangle EHB. Similarly, it may be proved, that the triangle GLK is equal to the triangle KFD,

therefore the whole triangle KGH is equal to the two triangles

EHB, KFD;

and consequently EG, FG bisect the rectangle ABCD.

I.

1. IN a given circle, place a straight line equal and parallel to a given straight line not greater than the diameter of the circle.

2. Trisect a given circle by dividing it into three equal sectors. 3. The centers of the circle inscribed in, and circumscribed about an equilateral triangle coincide; and the diameter of one is twice the diameter of the other.

4. If a line be drawn from the vertex of an equilateral triangle, perpendicular to the base, and intersecting a line drawn from either of the angles at the base perpendicular to the opposite side; the distance from the vertex to the point of intersection, shall be equal to the radius of the circumscribing circle.

5. If an equilateral triangle be inscribed in a circle, and a straight line be drawn from the vertical angle to meet the circumference, it will be equal to the sum or difference of the straight lines drawn from the extremities of the base to the point where the line meets the circumference, according as the line does or does not cut the base.

6. The perpendicular from the vertex on the base of an equilateral triangle, is equal to the side of an equilateral triangle inscribed in a circle whose diameter is the base. Required proof.

7. If an equilateral triangle be inscribed in a circle, and the adjacent arcs cut off by two of its sides be bisected, the line joining the points of bisection shall be trisected by the sides.

8. If an equilateral triangle be inscribed in a circle, any of its sides will cut off one-fourth part of the diameter drawn through the opposite angle.

9. The perimeter of an equilateral triangle inscribed in a circle is greater than the perimeter of any other isosceles triangle inscribed in the same circle.

10. If any two consecutive sides of a hexagon inscribed in a circle be respectively parallel to their opposite sides, the remaining sides are parallel to each other.

11. Prove that the area of a regular hexagon is greater than that of an equilateral triangle of the same perimeter.

12. If two equilateral triangles be inscribed in a circle so as to have the sides of one parallel to the sides of the other, the figure common to both will be a regular hexagon, whose area and perimeter will be equal to the remainder of the area and perimeter of the two triangles.

13. Determine the distance between the opposite sides of an equilateral and equiangular hexagon inscribed in a circle.

14. Inscribe a regular hexagon in a given equilateral triangle. 15. To inscribe a regular duodecagon in a given circle, and shew that its area is equal to the square on the side of an equilateral triangle inscribed in the circle.

II.

16. Describe a circle touching three straight lines.

17. Any number of triangles having the same base and the same vertical angle, will be circumscribed by one circle.

18. Find a point in a triangle from which two straight lines

drawn to the extremities of the base shall contain an angle equal to twice the vertical angle of the triangle. Within what limitations is this possible?

19. Given the base of a triangle, and the point from which the perpendiculars on its three sides are equal; construct the triangle. To what limitation is the position of this point subject in order that the triangle may lie on the same side of the base?

20. From any point B in the radius CA of a given circle whose center is C, a straight line is drawn at right angles to CA meeting the circumference in D; the circle described round the triangle ČBD touches the given circle in D.

21. If a circle be described about a triangle ABC, and perpendiculars be let fall from the angular points A, B, C, on the opposite sides, and produced to meet the circle in D, E, F, respectively, the circumferences EF, FD, DE, are bisected in the points A, B, Č.

22. If from the angles of a triangle, lines be drawn to the points where the inscribed circle touches the sides; these lines shall intersect in the same point.

23. The straight line which bisects any angle of a triangle inscribed in a circle, cuts the circumference in a point which is equidistant from the extremities of the side opposite to the bisected angle, and from the center of a circle inscribed in the triangle.

24. Let three perpendiculars from the angles of a triangle ABC on the opposite sides meet in P, a circle described so as to pass through P and any two of the points A, B, C, is equal to the circumscribing circle of the triangle.

25. If perpendiculars Aa, Bb, Cc be drawn from the angular points of a triangle ABC upon the opposite sides, shew that they will bisect the angles of the triangle abc, and thence prove that the perimeter of abe will be less than that of any other triangle which can

be inscribed in ABC.

26. Find the least triangle which can be circumscribed about a given circle.

27. If ABC be a plane triangle, GCF its circumscribing circle, and GEF a diameter perpendicular to the base AB, then if CF be joined, the angle GFC is equal to half the difference of the angles at the base of the triangle.

28. The line joining the centers of the inscribed and circumscribed circles of a triangle, subtends at any one of the angular points an angle equal to the semi-difference of the other two angles.

III.

29. The locus of the centers of the circles, which are inscribed in all right-angled triangles on the same hypotenuse, is the quadrant described on the hypotenuse.

30. The center of the circle which touches the two semicircles described on the sides of a right-angled triangle is the middle point of the hypotenuse.

31. If a circle be inscribed in a right-angled triangle, the excess of the sides containing the right angle above the hypotenuse is equal to the diameter of the inscribed circle.

32. Having given the hypotenuse of a right-angled triangle, and the radius of the inscribed circle, to construct the triangle.

33. ABC is a triangle inscribed in a circle, the line joining the middle points of the arcs AB, AC, will cut off equal portions of the two contiguous sides measured from the angle A.

IV.

34. Having given the vertical angle of a triangle, and the radii of the inscribed and circumscribed circles, to construct the triangle.

35. Given the base and vertical angle of a triangle, and also the radius of the inscribed circle, required to construct it.

36. Given the three angles of a triangle, and the radius of the inscribed circle, to construct the triangle.

37. If the base and vertical angle of a plane triangle be given, prove that the locus of the centers of the inscribed circle is a circle, and find its position and magnitude.

ཕ.

38. In a given triangle inscribe a parallelogram which shall be equal to one-half the triangle. Is there any limit to the number of such parallelograms?

39. In a given triangle to inscribe a triangle, the sides of which shall be parallel to the sides of a given triangle.

40. If any number of parallelograms be inscribed in a given parallelogram, the diameters of all the figures shall cut one another in the same point.

41. A square is inscribed in another, the difference of the areas is twice the rectangle contained by the segments of the side which are made at the angular point of the inscribed square.

42. Inscribe an equilateral triangle in a square, (1) When the vertex of the triangle is in an angle of the square. (2) When the vertex of the triangle is in the point of bisection of a side of the square. 43. On a given straight line describe an equilateral and equiangular octagon.

VI. 44. Inscribe a circle in a rhombus.

45. Having given the distances of the centers of two equal circles which cut one another, inscribe a square in the space included between the two circumferences.

46. The square inscribed in a circle is equal to half the square described about the same circle.

47. The square is greater than any oblong inscribed in the same circle.

48. A circle having a square inscribed in it being given, to find a circle in which a regular octagon of a perimeter equal to that of the square, may be inscribed.

49. Describe a circle about a figure formed by constructing an equilateral triangle upon the base of an isosceles triangle, the vertical angle of which is four times the angle at the base.

50. A regular octagon inscribed in a circle is equal to the rectangle

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