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GEOMETRICAL EXERCISES ON BOOK II.

PROPOSITION I. PROBLEM.

Divide a given straight line into two parts such, that their rectangle may be equal to a given square ; and determine the greatest square which the rectangle can equal.

Let AB be the given straight line, and let M be the side of the given square.

It is required to divide the line AB into two parts, so that the rectangle contained by them may be equal to the square on M.

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in D;

Bisect AB in C, with center C, and radius CA or CB, describe the semicircle ADB.

At the point B draw BE at right angles to AB and equal to M. Through E, draw ED parallel to AB and cutting the semicircle

and draw DF parallel to EB meeting AB in F. Then AB is divided in F, so that the rectangle AF, FB is equal to the square on M. (11. 14.) The square will be the greatest, when ED touches the semicircle,

or when M is equal to half of the given line AB.

PROPOSITION II. THEOREM. The square on the excess of one straight line above another is less than the squares on the two lines by twice their rectangle.

Let AB, BC be the two straight lines, whose difference is AC.

Then the square on AC is less than the squares on AB and BC by twice the rectangle contained by AB and BC.

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D

F E
Constructing as in Prop. 4. Book II.
Because the complement AG is equal to GE,

add to each CK,
therefore the whole AK is equal to the whole CE;

and AK, CE together are double of AK; but AK, CE are the gnomon AKF and CK, and AK is the rectangle contained by AB, BC;

therefore the gnomon AKF and CK

are equal to twice the rectangle AB, BC,
but AE, ČK are equal to the squares on AB, BC;

taking the former equals from these equals, therefore the difference of and the gnomon AKF is equal to the difference between the squares on AB, BC, and twice the rectangle AB, BC; but the difference AE and the gnomon AKF is the figure HF

which is equal to the square on AC. Wherefore the square on AC is equal to the difference between the squares on AB, BC, and twice the rectangle AB, BC.

PROPOSITION III. THEOREM.

In any triangle the squares on the two sides are together double of the squares on half the base and on the straight line joining its bisection with the opposite angle.

Let ABC be a triangle, and AD the line drawn from the vertex A to the bisection D of the base BC.

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From A draw AE perpendicular to BC. Then, in the obtuse-angled triangle ABD, (II. 12.); the square on AB exceeds the squares on AD, DB, by twice the rectangle BD, DE:

and in the acute-angled triangle ADC, (II. 13.); the square on AC is less than the squares on AD, DC, by twice

the rectangle CD, DE: wherefore, since the rectangle BD, DE is equal to the rectangle CD,

DE; it follows that the squares on AB, AC are double of the squares on AD, DB.

PROPOSITION IV. THEOREM.

If straight lines be drawn from each angle of a triangle bisecting the opposite sides, four times the sum of the squares on these lines is equal to three times the sum of the squares on the sides of the triangle.

Let ABC be any triangle, and let AD, BE, CF be drawn from A, B, C, to D, E, F, the bisections of the opposite sides of the triangle: draw AG perpendicular to BC.

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Then the square on AB is equal to the squares on BD, DA together with twice the rectangle BD, DG, (11. 12.)

and the square on AC is equal to the squares on CD, DA dimi

nished by twice the rectangle CD, ; (11. 13.) therefore the squares on AB, AC are equal to twice the square on

BD, and twice the square on A.D; for DC is equal to BD: and twice the squares on AB, AC

are equal to the square on BC, and four times the square on AD: for BC is twice BD. Similarly, twice the squares on AB, BC are equal to the square on AC, and four times the square on BE: also twice the squares on BC, CA are equal to the square on AB, and four times the square on FC:

hence, by adding these equals, four times the squares on AB, AC, BC are equal to four times the squares on AD, BE, CFtogether with the squares on AB, AC, BC:

and taking the squares on AB, AC, BC from these equals, therefore three times the squares on AB, AC, BC are equal to four times the squares on AD, BE, CF.

PROPOSITION V. THEOREM.

The sum of the perpendiculars let fall from any point within an equilateral triangle, will be equal to the perpendicular let fall from one of its angles upon the opposite side. Is this proposition true when the point is in one of the sides of the triangle? In what manner must the proposition be enunciated when the point is without the triangle ?

Let ABC be an equilateral triangle, and P any point within it: and from Pletfall PD, PE, PF perpendiculars on the sides AB, BC, CA respectively, also from A let fall ÅG perpendicular on the base BC

Then AG is equal to the sum of PD, PE, PF.

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From P draw PA, PB, PC to the angles A, B, C. Then the triangle ABC is equal to the three triangles PAB, PBC, PCA.

But since every rectangle is double of a triangle of the same base and altitude, (1. 41.)

therefore the rectangle AG, BC, is equal to the three rectangles AB, PD; AC, PF and BC, PE.

Whence the line AG is equal to the sum of the lines PD, PE, PF. If the point P fall on one side of the triangle, or coincide with E:

then the triangle ABC'is equal to the two triangles APC, BPA: whence AG is equal to the sum of the two perpendiculars PD, PF. If the point P fall without the base B C of the triangle :

then the triangle ABC is equal to the difference between the sum of the two triangles APC, BPA, and the triangle PCB.

Whence AG is equal to the difference between the sum of PD, PF, and PE.

I. 6. If the straight line AB be divided into two unequal parts in D, and into two unequal parts in E, the rectangle contained by AE, EB, will be greater or less than the rectangle contained by AD, DB, according as E is nearer to, or further from, the middle point of AB, than D.

7. Produce a given straight line in such a manner that the square on the whole line thus produced, shall be equal to twice the square on the given line.

8. If AB be the line so divided in the points C and E, (fig. Euc. II. 5.) Shew that AB=4.CD? + 4. AD.DB.

9. Divide a straight line into two parts, such that the sum of their squares may be the least possible.

10. Divide a line into two parts, such that the sum of their squares shall be double the square on another line.

11. Shew that the difference between the squares on the two unequal parts (fig. Euc. II. 9.) is equal to twice the rectangle contained by the whole line, and the part between the points of section.

12. Shew how in all the possible cases, a straight line may be geometrically divided into two such parts, that the sum of their squares shall be equal to a given square.

13. Divide a given straight line into two parts, such that the squares on the whole line and on one of the parts shall be equal to twice the square on the other part.

14. Any rectangle is the half of the rectangle contained by the diameters of the squares on its two sides.

15. If a straight line be divided into two equal and into two unequal parts, the squares on the two unequal parts are equal to twice the rectangle contained by the two unequal parts, together with four times the square on the line between the points of section.

16. If the points C, D be equidistant from the extremities of the straight line AB, shew that the squares constructed on AD and AC, exceed twice the rectangle AC, AD by the square constructed on CD.

17. If any point be taken in the plane of a parallelogram from which perpendiculars are let fall on the diagonal, and on the sides which include it, the rectangle of the diagonal and the perpendicular on it, is equal to the sum or difference of the rectangles of the sides and the perpendiculars on them.

18. ABCD is a rectangular parallelogram, of which A, C are opposite angles, E any point in BC, F any point in CD. Prove that twice the area of the triangle AEF together with the rectangle BE, DF is equal to the parallelogram AC.

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19. Shew how to produce a given line, so that the rectangle contained by the whole line thus produced, and the produced part, shall be equal to the square (1) on the given line (2) on the part produced.

20. If in the figure Euc. 11. 11, we join BF and CH, and produce CH to meet BF in L, CL is perpendicular to BF

21. If a line be divided, as in Euc. II. 11, the squares on the whole line and one of the parts are together three times the square on the

22. If in the fig. Euc. II. 11, the points F, D be joined cutting LAAB, GHK in f, d respectively; then shall Ff = Dd.

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jother part.

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23. If from the three angles of a triangle, lines be drawn to the points of bisection of the opposite sides, the squares on the distances between the angles and the common intersection, are together one-third of the squares on the sides of the triangle.

24. ABC is a triangle of which the angle at C is obtuse, and the angle at B is half a right angle: D is the middle point of AB, and CE is drawn perpendicular to AB. Shew that the square on A C'is double of the squares on AD and DE.

25. If an angle of a triangle be two-thirds of two right angles, shew that the square on the side subtending that angle is equal to the squares on the sides containing it, together with the rectangle contained by those sides.

26. The square described on a straight line drawn from one of the angles at the base of a triangle to the middle point of the opposite side, is equal to the sum or difference of the square on half the side bisected, and the rectangle contained between the base and that part of it, or of it produced, which is intercepted between the same angle and a perpendicular drawn from the vertex.

27. ABC is a triangle of which the angle at C is obtuse, and the angle at B is half a right angle: D is the middle point of AB, and ÇE is drawn perpendicular to AB. Shew that the square on AC is double of the squares on AD and DE.

28. Produce one side of a scalene triangle, so that the rectangle under it and the produced part may be equal to the difference of the squares on the other two sides.

29. Given the base of any triangle, the area, and the line bisecting the base, construct the triangle.

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