Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

or AB2

Therefore the squares on BD and DA are equal to the ..BD2+DA2 squares on BC, CD, DA, and twice the rectangle BC, CD. =BC2+ But the square on BA is equal to the squares on BD and (CD3+ DA, because the angle at D is a right angle (I. 47); And the square on CA is equal to the squares on CD and BC2+ DA (I. 47);

Therefore the square on BA is equal to the squares on BC and CA, and twice the rectangle BC, CD; that is, the square on BA is greater than the squares on BC and CA by twice the rectangle BC, CD.

Therefore, in obtuse-angled triangles, &c. Q.E.D.

Proposition 13.-Theorem.

In every triangle, the square on the side subtending an acute angle is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle and the acute angle.

Let ABC be any triangle, and the angle at B an acute angle; and on BC, one of the sides containing it, let fall the perpendicular AD from the opposite angle (I. 12).

DA2)

+2BC CD

AC2
+2BC CD.

+AB2.

The square on AC, opposite to the angle B, shall be less AC2=CB2 than the squares on CB and BA, by twice the rectangle 2CB BD. CB, BD.

CASE I. First, let AD fall within the triangle ABC.

Δ

PROOF.-Because the straight line CB is divided into two parts in the point D, The squares on CB and BD are equal to twice the rectangle contained by CB, B BD, and the square on DC (II. 7).

To each of these equals add the square on DA.

For

CB2+BD2 =2CB BD

Č +DC2.

Therefore the squares on CB, BD, DA are equal to twice.CB2+

the rectangle CB, BD, and the squares on AD and DC.

(BD2+

DA2)

+(AD2+ DC2),

But the square on AB is equal to the squares on BD and =2CB BD DA, because the angle BDA is a right angle (I. 47); And the square on AC is equal to the squares on AD and DC (I. 47);

Therefore the squares on CB and BA are equal to the

[ocr errors]

CB2+BA2 =2CB.DB

+AC2.

..AC =CB2

+ BA2 -: 2CB BD.

AB2 AC2 +CB2+ BC CD.

...AB2+
BC2
=AC2+
2(BC2+
BC CD).

Now

DB BC=

square on AC, and twice the rectangle CB, BD; that is, the square on AC alone is less than the squares on CB and BA by twice the rectangle CB, BD.

B

CASE II.

Secondly, let AD fall without the triangle ABC. PROOF.-Because the angle at D is a right angle (Const.), the angle ACB is greater than a right angle (I. 16);

C

Therefore the square on AB is equal to the squares on AC and CB, and twice the rectangle BC, CD (II. 12).

To each of these equals add the square on BC.

Therefore the squares on AB and BC are equal to the square on AC, and twice the square on BC, and twice the rectangle BC, CD (Ax. 2).

But because BD is divided into two parts at C,

The rectangle DB, BC is equal to the rectangle BC, CD and the square on BC (II. 3);

And the doubles of these are equal, that is, twice the BCCD rectangle DB, BC is equal to twice the rectangle BC, CD and twice the square on BC;

BC2.

...AC2 =AB2+ BC22BC.DB.

Therefore the squares on AB and BC are equal to the square on AC, and twice the rectangle DB, BC; that is, the A square on AC alone is less than the squares on AB and BC by twice the rectangle DB, BC.

B

CASE III. Lastly, let the side AC be perpendicular to BC.

PROOF. Then BC is the straight line between the perpendicular and the acute angle at B; and it is manifest that the squares on AB and BC are equal to the square on AC, and twice the square on BC (I. 47, and Ax. 2).

Therefore, in every triangle, &c. Q.E.D.

Proposition 14.-Problem.

To describe a square that shall be equal to a given rectilineal figure.

Let A be the given rectilineal figure.

It is required to describe a square that shall be equal to A.

CONSTRUCTION.-Describe the rectangular parallelogram BCDE equal to the rectilineal figure A. (I. 45).

If then the sides of it, BE, ED, are equal to one another it is a square, and what

was required is now

done.

But if they are not equal, produce one of them, BE, to F, and make EF equal to ED (I. 3).

[blocks in formation]

Bisect BF in G (I. 10), and from the centre G, at the distance GB, or GF, describe the semicircle BHF;

Produce DE to H, and join GH;

Then the square described upon EH shall be equal to the rectilineal figure A.

PROOF.-Because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in the point E,

The rectangle BE,EF, together with the square on GE, BEEF is equal to the square on GF (II. 5).

But GF is equal to GH;

+GE2 =GF2

=GH2 =GE2+

Therefore the rectangle BE,EF, together with the square EH2.

on GE, is equal to the square on GH.

But the square on GH is equal to the squares on GE and EH (I. 47);

Therefore the rectangle BE,EF, together with the square ..BE EF on GE, is equal to the squares on GE and EH.

Take away the square on GE, which is common to both; Therefore the rectangle BE,EF is equal to the square on EH (Ax. 3).

But the rectangle contained by BE and EF is the parallelogram BD, because EF is equal to ED (Const.); Therefore BD is equal to the square on EH.

or BD =EH2.

But BD is equal to the rectilineal figure A (Const.); Therefore the square on EH is equal to the rectilineal Hence figure A.

Therefore, a square has been made equal to the given rectilineal figure A, viz., the square described on EH. Q.E.F.

EH2=A.

EXERCISES ON BOOK II.

PROP. 1-11.

1. Divide a given straight line into two such parts that the rectangle contained by them may be the greatest possible.

2. The sum of the squares of two straight lines is never less than twice the rectangle contained by the straight lines.

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

4. Given the sum of two straight lines and the difference of their squares, to find the lines.

5. In any triangle the difference of the squares of the sides is equal to the rectangle contained by the sum and difference of the parts into which the base is divided by a perpendicular from the vertical angle.

6. Divide a given straight line into such parts that the sum of their squares may be equal to a given square.

7. If ABCD be any rectangle, A and C being opposite angles, and O any point either within or without the rectangle-OA2 + OC2 OB2+ OD2.

8. Let the straight line AB be divided into any two parts in the point C. Bisect CB in D, and take a point E in AC such that EC CD. Then shall AD2 = AE2 + AC CB.

9. If a point C be taken in AB, and AB be produced to D so that BD and AC are equal, show that the squares described upon AD and AC together exceed the square upon AB by twice the rectangle contained by AE and AC.

10. From the hypothenuse of a right-angled triangle portions are cut off equal to the adjacent sides. Show that the square on the middle segment is equal to twice the rectangle under the extreme segments.

11. If a straight line be divided into any number of parts, the square of the whole line is equal to the sum of the squares of the parts, together with twice the rectangles of the parts taken two and two together.

12. If ABC be an isosceles triangle, and DE be drawn parallel to the base BC, cutting in D and E either the side or sides produced, and EB be joined; prove that BE2 BC DECE2.

=

PROP. 12-14.

13. In any triangle show that the sum of the squares on the sides is equal to twice the square on half the base, and twice the square on the line drawn from the vertex to the middle of the base.

14. If squares are described on the sides of any triangle, find the difference between the sum of two of the squares and the third square, and show from your result what this becomes when the angle opposite the third square is a right angle.

15. Show also what the difference becomes when the vertex of the triangle is depressed until it coincide with the base.

16. The square on any straight line drawn from the vertex of an isosceles triangle, together with the rectangle contained by the segments of the base, is equal to the square upon a side of the triangle.

17. If a side of a triangle be bisected, and a perpendicular drawn from the middle point of the base to meet the side, then the square of the altitude of the triangle exceeds the square upon half the base by twice the rectangle contained by the side and the straight line between the points of section of the side.

18. In any triangle ABC, if perpendiculars be drawn from each of the angles upon the opposite sides, or opposite sides produced, meeting them respectively in D, E, F, show that

[ocr errors]

BA2 + AC2 + CB2 2 AE AC + 2 CDCB + 2 BF BA;

all lines being measured in the same direction round the triangle. 19. Construct a square equal to the sum of the areas of two given rectilinear figures.

20. The base of a triangle is 63 ft., and the sides 25 ft. and 52 ft. respectively. Show that the segments of the base, made by a perpendicular from the vertex, are 15 ft. and 48 ft. respectively, and that the area of the triangle is 630 sq. ft.

21. In the same triangle, show that the length of the line joining the vertex with the middle of the base is 22.9 ft.

22. A ladder, 45 ft. long, reaches to a certain height against a wall, but, when turned over without moving the foot, must be shortened 6 ft. in order to reach the same height on the opposite side. Supposing the width of the street to be 42 ft., show that the height to which the ladder reaches is 36 ft.

23. The base and altitude of a triangle are 8 in. and 9 in. respectively; show that its area is equal to a square whose side is 6 in. Prove your result by construction.

24. On the supposition that lines can be always expressed exactly in terms of some unit of length, what geometrical propositions may be deduced from the following algebraical identities?—

[blocks in formation]
« ΠροηγούμενηΣυνέχεια »