Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Book II.

PROP. XI. PROB.

To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part.

Let AB be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

a

b

Upon AB describe the square ABDC; bisect AC in E, and join BE; produce CA to F, and make EF equal to EB, and upon AF describe the square FGHA; AB is divided in H, so that the rectangle AB, BH is equal to the square of AH.

e

G

d

a 46. 1.

b 10. 1.

c 3.1.

d

e 47. 1.

Produce GH to K: Because the straight line AC is bisected in E, and produced to the point F, the rectangle CF.FA, together with the square of AE, is equal à 6, 2. to the square of EF: But EF is equal to EB: therefore the rectangle CF.FA, together with the square of AE, is equal to the square of EB. And the squares of BA, AE, are equal to the square of EB, because the angle EAB is F a right angle; therefore the rectangle CF.FA, together with the square of AE, is equal to the squares of BA, AE: take away the square of AE, A which is common to both, therefore the remaining rectangle CF.FA is equal to the square of AB. Now the figure FK is the rectangle CF.FA, for AF is equal to FG; and AD is the square of AB; therefore FK is equal to AD: take away the common

E

C

K

H

B

D

part AK, and the remainder FH is equal to the remainder HD. But HD is the rectangle AB.BH, for AB is

Book II. equal to BD; and FH is the square of AH; therefore the rectangle AB.BH is equal to the square of AH: Wherefore the straight line AB is divided in H, so that the rectangle AB.BH is equal to the square of AH. Which was to be done.

a 12. 1.

b 4. 2.

c 47. 1.

PROP. XII. THEOR.

In obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted between the perpendicular and the obtuse angle.

Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn perpendicular to BC produced: The square of AB is greater than the squares of AC, CB, by twice the rectangle BC.CD.

Because the straight line BD is divided into two parts in the point C, BD2 = b BC2 + CD2 + 2BC.CD; add AD2 to both: Then BD2 + AD2 = BC2 + CD2 + AD + 2BC.CD. But AB2 BD2 + AD2, and AC2CD2+ AD2c; therefore, AB2 BC2 + AC2 + 2BC.CD; that is, AB is B greater than BC2 + AC2 by

C

2BC.CD. Therefore, in obtuse angled triangles, &c.

Q. E. D.

Book II.

PROP. XIII. THEOR.

In every triangle, the square of the side subtending any of the acute angles, is less than the squares of 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 upon it from the opposite angle, and the acute angle.

[ocr errors]

Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular AD from the opposite angle: The square of AC, opposite to the angle B, is less than the squares of CB, BA by twice the rectangle CB.BD.

A

First, Let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D, BC2 + BD2 = 2BC.BD+CD2. Add to each AD2; then BC2 + BD2 + AD2 2BC.BD + CD2 + AD2. But BD2 + AD2 AB2, and CD2 + DA2 = AC2; therefore BC2 + AB2 -2 BC.BD + AC; that is,

B

AC2 is less than BC2+ AB2 by 2BC.BD.

Secondly, Let AD fall without the triangle ABC *: Then because the angle at D is a right angle, the angle ACB is greater than a right angle, and AB

e

a 12.1.

b 7. 2.

c 47. 1.

AC d 16. 1.

+ BC2+2BC.CD. Add BC2 to each; then AB2 + BC2 e 12. 2.) AC2+2BC2 + 2BC.CD. But because BD is divided

into two parts in C, BC2+BC.CD=f BC.BD, and 2BC2 £3. 2. +2BC.CD=2BC.BD: therefore AB2 + BC2 = AC2+ 2 BC.BD; or AC2 is less than AB2 + BC2 by 2BD.BC.

See figure of the last Proposition.

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

a 45. 1.

PROP. XIV. PROB.

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.

Describe the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it, BE, ED are equal to one another, it is a square, and what was required is done; but if they are not equal, produce one of them, BE to F, and make EF equal to ED, and bisect BF in G: and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H, and join GH. Therefore, because the straight line BF is divided into two equal parts in G, and into two unequal parts in E, the rectangle BE.EF, b 5. 2. together with the square of EG, is equal to the square of GF: but GF is equal to GH; therefore the rectangle BE.EF, together with the square of EG, is equal to the square of GH: But the squares of HE and EG are € 47. 1. equal to the

c

square of GH:

Therefore also the rectangle BE.EF together with the square of EG, is equal to the squares of HE and EG. Take

[blocks in formation]

away the square of EG, which is common to both, and the Book FI. remaining rectangle BE. EF is equal to the square of EH. But BD is the rectangle contained by BE and EF, because EF is equal to ED; therefore BD is equal to the square of EH; and BD is also equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. Which was to be done.

PROP. A. THEOR.

If one side of a triangle be bisected, the sum of the See N. squares of the other two sides is double of the square of half the side bisected, and of the square of the line drawn from the point of bisection to the opposite angle of the triangle.

Let ABC be a triangle, of which the side BC is bisected in D, and DA drawn to the opposite angle; the squares of BA and AC are together double of the squares of BD and DA.

From A draw AE perpendicular to BC, and because BEA is a right angle, AB

BE+AE and AC2=

CE+AE2; wherefore AB2

+ AC2 BE2 + CE2 + pige site 2AE2. But because the line

BC is cut equally in D, and unequally in E, BE2 + CE2 =b2BD2 + 2DE2; there

fore AB2+AC2

2DE2+2AE2.

2BD2 +B

a

D

a 47.

[merged small][ocr errors]

Now, DE2+ AE AD2, and 2DE2 + 2AE2 = 2AD2; wherefore AB2+ AC2 = 2BD2 + 2AD2. Therefore, &c. Q. E. D.

[ocr errors][ocr errors]
« ΠροηγούμενηΣυνέχεια »