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THE

ELEMENTS

OF EUCLID.

BOOK II.

DEFINITIONS.

Α Ε

D

1. Every right-angled parallelogram, or rectangle, is said to

be contained by any two of the straight lines which contain

one of the right angles. II. In every parallelogram, any of the parallelograms about

a diameter, together with the two com-
plements, is called a Gnomon. • Thus
'the parallelogram Hg, together with
• the complements AF, FC, is the gno- H
'mon, which is more briefly expressed

BG ' by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which • make the gnomon.'

K C

PROPOSITION I. THEOREM.--If there be two straight lines, one of which is di

vided into any number of parts, the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Let A and BC be two straight lines ; and let BC be divided into any parts in the points D, E: the rectangle con

# 11. 1.

3. 1.

31. 1. # 31.1.

А

tained by the straight lines A, BC, shall be equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC.

From the point B, draw * BF at right B D EC
angles to BC, and make BG equal * to A;
and through G, draw * GH parallel to

G
BC; and through D, E, C, draw * DK,

K L Η
EL, CH parallel to BG: then the rectangle F
BH is equal to the rectangles BK, DL,
EH: but BH is contained by A, BC, for it is contained by
GB, BC, and GB is equal + to A; and BK is contained by A,
BD, for it is contained by GB, BD, of which GB is equal to
A; and DL is contained by A, DE, because DK, that is * BG,
is equal to A ; and in like manner the rectangle Eh is con-
tained by A, EC: therefore the rectangle contained by A,
BC, is equal to the several rectangles contained by A, BD,
and by A, DE, and by A, EC. Wherefore, if there be two
straight lines, &c. Q. E. D.

+ Constr.

* 34. 1.

PROPOSITION II.

Theor.-If a straight line be divided into any two parts, the

rectangles contained by the whole and each of the parts, are together equal to the square of the whole line.

46. 1. 31. 1.

Let the straight line AB be divided into any two parts in the point C: the rectangle contained by AB, BC, together with the rectangle # AB, AC, shall be equal to the square of AB.

Upon AB describe* the square ADEB; and A CB through c, draw * CF, parallel to AD or BE: then AE is equal to the rectangles AF, CE: but AE is the square of AB; and AF is the rectangle contained by BA, AC, for it is con

FE tained by DA, AC, of which AD is equal + to AB; and CE is contained by AB, BC, for BE is equal to AB: therefore the rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. If, therefore, a straight line, &c.

D

+ 30 Def.

Q. E, D.

# N.B.---To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC.

PROPOSITION III.

. 46. 1.

THEOR.-If a straight line be divided into any two parts, the

rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

Let the straight line AB be divided into any two parts in the point C: the rectangle AB, BC shall be equal to the rectangle AC, CB, together with the square of BC.

Upon BC, describe * the square CDEB, and produce ED to F; and through A, с

B draw * AF parallel to CD or BE: then the rectangle AE is equal to the rectangles AD, CE: but AE is the rectangle contained by AB, BC, for it is contained by F D E AB, BE, of which BE is equal t to BC; and AD is contained + 30 Def. by AC, CB, for CD is equal to CB; and DB is the square

of BC: therefore the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. If, therefore, a straight line, &c.

Q. E. D.

A

• 31. 1.

PROPOSITION IV.

Theor.-If a straight line be divided into any two parts, the

square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in C: the square of AB shall be equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB.

Upon AB, describe * the square ADEB, and join BD; and * 46. I. through c, draw * CGF parallel to AD or BE, and through

* 31. 1. G, draw UK parallel to AB or DE. And be

A C B cause CF is parallel + to AD, and BD falls

+ Constr. upon them, the exterior angle BGC is equal HI

K

* 29. 1. to the interior and opposite angle ADB; but ADB is equal * to the angle ABD, because

D F E

5. I. BA is equal to AD +, being sides of a square ;

+ 30 Def. wherefore the angle CGB is equal † to the angle CBG; and + 1 As. therefore the side BC is equal * to the side CG; but CB is * 6. 1. equal * also to GK, and co to BK; wherefore the figure . 34. 1.

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CGKB is † equilateral : it is likewise rectangular; for, since CG is parallel to BK, and CB meets them, therefore the angles KBC, GCB are equal + to two right angles: but KBC is a t right angle ; wherefore GCB is at right angle: and therefore also the angles * CGK, GKB, opposite to these are right angles ; and therefore CGKB is rectangular ; but it is also equilateral, as was demonstrated ; wherefore it is a t square, and it is upon the side CB : for the same reason HF also is a square, and it is upon the side HG, which is equal t to AC: therefore HF, CK are the squares of AC, CB: and because the complement AG is equal* to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal + to CB, therefore GE is also equal + to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB ; and HF, CK are the squares of AC, CB ; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB: but HF, CK, AG, GE make

up the whole figure ADEB, which is the square of AB : therefore the square of AB is equal + to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore, if a straight line, &c.

Cor. From the demonstration it is manifest, that parallelograms about the diameter of a

square are likewise

+ 1 Ax.

Q. E. D.

squares.

PROPOSITION V.

Theor.-If a straight line be divided into two equal parts and

also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D: the rectangle AD, DB, together with the square of CD, shall be equal to the square of CB. Upon CB, describe * the square CEFB,

DB join BE: and through D, draw * DHG

M

K
parallel to CE or BF; and through H,
draw KLM parallel to CB or EF; and

E GF
also through A, draw AK parallel to
CL or BM. And because the complement CH is equal * to
the complement HF, to each of these add DM; therefore the

# 46. 1. * 31. 1.

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* 43. 1.

• 36. 1.

and 34. 1.

whole CM is equal t to the whole DF: but CM is equal * † 2 Ax. to AL, because AC is equal + to CB; therefore also AL is

+ Hyp. equal + to DF: to each of these add ch, and the whole + i Ax. AH is equal + to DF and ch: but AH is the rectangle † 2 Ax. contained by AD, DB, for DH is equal * to DB; and DF to- * Cor. 4. 2. gether with ch is the

& 30 Def. gnomon CMG ; therefore the gnomon CMG is equal t to the rectangle AD, DB: to each of these add † 1 Ax. LG, which is equal * to the square of CD; therefore the * Cor. 4. 2. gnomon CMG, together with LG is equal + to the rectangle

† 2 Ax. AD, DB, together with the square of CD: but the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB ; therefore the rectangle AD, DB, together with the

square of CD, is equal to the square of CB. Wherefore, if a straight line, &c.

From this proposition it is manifest, that the difference of the squares of two unequal lines, AC, CD, is equal to the rectangle contained by their sum and difference.

Q. E. D.

PROPOSITION VI.!

THEOR.-If a straight line be bisected, and produced to any

point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced.

46. 1. * 31. 1.

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E

Let the straight line AB be bisected in C, and produced to the point D: the rectangle AD, DB, together with the square of CB, shall be equal to the square of CD.

Upon CD, describe* the square CEFD, A с B D join DE; and through B, draw * BHG L

K

M parallel to CE or DF; and through H, draw KLM parallel to AD or EF; and also through A, draw AK parallel to CL

G F or DM. And because AC is equal + to

+ Hyp: CB, the rectangle AL is equal * to CH; but CH is equal * to HF; therefore also Al is equal t to HF: to each of these add

+1 Ax. CM; therefore the whole AM is equal t to the gnomon CMG: * 2 Ax. but AM is the rectangle contained by AD, DB, for DM is equal * to DB : therefore the gnomon CMG is equal + to the

Cor. 4, 2.

& 30 Def. rectangle AD, DB : add to each of these LG, which is equal +

f 1 Ax. to the square of CB ; therefore the rectangle AD, DB, to- † Cor. 4. 2.

and 34. 1.

36. 1.

• 43. 1.

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