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Let A and BC be two straight lines; and let BC be divid- Book II. ed into any parts in the points D, E; the rectangle contained so by the straight lines A, BC is

B D E C 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 drawa BF G

a ll. 1. at right angles to BC, and

K L H make BG equal 6 to A; and

b 3. 1. through G draw GH parallel F

A

c 31. 1. to BC, and through D, E, C, drawC DK, EL, CH parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH; and 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 d 34. 1. A; and in like manner the rectangle EH is contained 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 also by A, EC. Wherefore, if there be two straight lines," &c. Q. E. D.

PROP. 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.

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

Upon A B describe the square ADEB, and through C draw CF, parallel to AD or BE; then AE is equal to the rectangles AF, CE; and AE is the square of AB; and AF is the rectangle contained by BA, D F E

• 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.

a 46. 1. b 31. 1.

Book II. AC; for it is contained 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. Q. E. D.

PROP. III. THEOR.

If a straight line be divided into any two parts, the rectangle contained ty the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the foresaid part.

a 46. 1.

b

b 31. 1.

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

Upon BC described the square CDEB, and produce A C B ED to F, and through A draw AF parallel to CD or BE; then the rectangle A E is equal to the rectangles AD, CE; and AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and AD is con- F D E tained 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.

PROP. 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 is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB.

b .

Upon AB describe a the square ADEB, and join BD, and Book II. through C draw • CGF parallel to AD or BE, and through G draw HK parallel to AB or DE: and because CF is paral- a 46. 1. lel to AD, and BD falls upon them, the exterior angle BGC is equal to the interior and opposite angle ADB; but ADB c 29. 1. is equal to the angle ABD, because BA is equal to AD, d 5. 1. being sides of a square; wherefore the angle CGB is equal to the angle А C B GBC; and therefore the side BC is equal e to the side CG: But CB is

e 6. 1.

G equal also to GK, and CG to BK; H

f 34. 1.

K wherefore the figure CGKB is equilateral: It is likewise rectangular; for CG is parallel to BK, and CB meets them; the angles KBC, GCB are therefore equal to two right D F E angles; and KBC is a right angle; wherefore GCB is a right angle; and therefore also the angles CGK, GKB opposite to these, are right angles, and CGKB is rectangular: But it is also equilateral, as was demonstrated; wherefore it is a 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 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 8 43. 1. 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. Q. E. D.

Cor. From the demonstration, it is manifest, that the parallelograms about the diameter of a square are likewise squares.

Book II.

PROP. 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, is equal to the square

of CB. a 46. 1. Upon CB describe a the square CEFB, join BE, and through b 31. 1. D drawb DHG parallel to CE or BF; and through H draw

KLM parallel to CB or EF; and also through A draw AK

parallel to CL or BM: And because the complement CH is c 43. 1. equal to the complement HF, to each of these add DM;

therefore the whole CM
is equal to the whole DF;

A С D B d 36. 1. but CM is equal d to AL, because AC is equal to

L H
CB; therefore also AL is K

M M
equal to DF. To each of
these add CH, and the
whole AH is equal to DF
and CH: But AH is the

E GF rectangle contained by

AD, DB, for DH is equal Cor. 4. 2.e to DB: and DF together with CH is the gnomon CMG;

therefore the gnomon CMG is equal to the rectangle AD, DB: To each of these add LG, which is equal to the square of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle 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. R.E.D.

Cor. 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 AD and difference DB.

OF EUCLID.

47

PROP. VI. THEOR.

Book II,

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.

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, is equal to the square of CD.

Upon CŨ describe a the square CEFD, join DE, and a 46. 1. through B drawb BHG parallel to CE or DF, and through b 31. I. H draw KLM parallel to AD or EF, and also through A draw AK parallel to CL or DM: and because A с B D AC is equal to CB, the rectangle AL is equal

L H

c 36. 1. to CH; but CH is K'

M equal to HF; therefore also AL is equal to HF: To each of these add CM; therefore the whole AM is equal to

E G F the gnomon CMG : and AM is the rectangle contained by AD, DB, for DM is equal e to DB: Therefore the gnomon CMG is equal to the e Cor. 4. 2. rectangle AD, DB: Add to each of these LG, which is equal to the square of CB, therefore the rectangle AD, DB, together with the square of CB, is equal to the gnomon CMG and the figure LG: But the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line,” &c. Q.E. D.

с

d 43. 1.

PROP. VII. THEOR.

If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part.

Let the straight line AB be divided into any two parts in

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