Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Book II.

PRO P. VI. T H E O R.

IF.
Fa right line be divided into equal parts, and another line add-

ed to it, the rectangle contained by the whole and added line as one side of the rectangle, and the added line for the other side, together with the square of half the line, are equal to the square of the half and added line, as one side of the square.

Let the right line AB be bisected in C, and BD added to it, the rectangle under AD, DB, together with the square of BC, are equal to the square of CD.

Describe the square CEFD; construct the figure ; and compleat the parallelogram under AC, CL.

Then the parallelograms AL, CH, are equal“; but CH is e. a 36. . qual to HF b; add BM to both; then CM is equal to BF; add b 43. 1. AL to both; then AM is equal to the gnomon CMG. To each add LG, that is, the square of CB C; then AM, LG, are c Cor. 4. equal to the gnomon CMG, and LG; that is, the rectangle under AD, DB, for DM is equal to DB, together with the square CB, are equal to the square of CD. Wherefore, &c.

PRO P. VII, . THEO R.

Fa right line be any how cut, the square of the whole line, and

one of the parts, is equal to twice the reEtangle contained by the whole line, and said part, together with the square of the o

[ocr errors]

ther part.

Let the right line AB be any how cut in C, the squares of AB, BC, are equal to twice the rectangle under AB, BC, and the square of AC. Upon AB describe the square ADEB“, and construct the fi- a 46.3.

b 43. 1. and gure; then, because the rectangle AF is equal to CE 6, and AF, CE, together, are equal to twice AF, that is, equal to the gnomon AFK, together with the square of CB, that is, CF <; add c Cor. 4: HK to both; then twice AF, and HK, are equal to the gnomon AFK, and the squares of AC, BC; that is, to the squares of AB, BC. Wherefore, &c,

Ax. 2. I.

PROPS

[ocr errors][merged small]

PRO P. VIII. THE O R.

[ocr errors]

a right line be cut into two parts, four times the rectangle

under the whole line, and one of the parts, together with the Square of the other part, are equal to the square of the whole line, and the first part taken as the side of the square.

Let the right line AB be any how cut in C, four times the rectangle under AB, BC, together with the square of AC, are equal to the square of AD; that is, AB produced to D, so that BD equal BC. Upon AD describe the square AEFD, and

construct the double figure. Then, because BN, GR, are a Cor.4. squares, and CK, BN, are equal parallelograms b; but the sides b 36, 1

CB, BK, are equal, and CBK is a right angle, for it is equal d 34: 1.and to BDN“; therefore CK is a squared. For the same reason, ko def. 30.1. is a square; therefore CK, BN, GR, KO, are each squares;

but they are constitute upon equal right lines ; therefore equal to one another, and, together, quadruple KC. But the rec

tangle AG is equal to MP, and PL to RF; but MP is equal f 43. 1. to PLf; therefore the four rectangles are quadruple AG;

and the four squares and four rectangles quadruple the rectangle AK, that is, the rectangle under AB, BC; add the square XH, that is, the square of AC; then four times AK, that is, four times the rectangle under AB, BC, together with the square of AC, are equal to the square of AD. Wherefore, &c.

PRO P. IX. THEO R.

IFS

a right line be cut into two equal parts, and into two unequal

parts, the squares of the two unequal parts are double the square of the half line, and double the square of the intermediate part.

Let the right line AB be cut equally in C, and unequally in D, the squares of AD, BD, are double the squares of AC, CD.

For, through C draw CE, at right angles, to AB, and equal to AC, or CB ; join EA, EB; through D draw DF parallel to to CE, and FG through F, parallel to AB; join AF.

Then, because AC is equal to CE, and the angle ACE a * C.F. 32. right angle, the angles AEC, EAC are each half right angles",

and the squares of AC, CE double the square of AC; but the b 47. 1. square of AE is equal to the squares of AC, CEb; therefore,

double the square of AC. For the same reason, the angles CEB, Boor II. EBC are each half right angles ; but the angle EGF is a right angle; therefore GFE is half a right angle; therefore the fides c 29. 1. EG, GF are equal d; but the square of EF is equal to the squares d.6.1. of EG, GF, or double the square of GF or CD °; but the squares e 34. 5. of AE, EF are equal to the square of AF, for the angle AEF is a right angle; but the squares of AE, FF are double the squares of AC, CD; therefore the square of AF is double the squares of AC, CD; but the angle DFB is half a right angle ; for it is equal to CEB °; therefore, DFB, DBF are each half right angles ; therefore FD, DB are equal; but the square of AF is equal to the squares of AD, DFb, or DB; therefore, the squares of AD, DB are double the squares of AC, CD. Wherefore, &c.

PRO P. X. THEO R.

IF F a right line be cut into two equal parts, and another right

line added to it, the square of the whole and added line taken as one line, and the square of the added line, are double the Square of the half line, and double the square of the half and added line, taken as one line.

Let the right line AB be bisected in C, and BD added to it, the squares of AD, DB are double the squares of AC, CD,

For, from the point C, draw CE perpendicular to AB, and equal to AC or CB ; join AE, EB ; through E, draw EF parallel to AD; and through D, draw DF parallel to CE.

Because AC is equal to CE, and the angle ACE a right angle, each of the angles AEC, EAC are half right angles; and therefore the square of AE is equal to the squares of AC, CE, or double the square of AC . For the same reason, CEB, CBE a 47. 1. are each half right angles; therefore AEB is a right angle; but the angles FEC, ECD are equal to two right angles b, and ECD b 29. 1° is a right angle; therefore CEF is likewise a right angle; therefore CF is a rectangle ; therefore, the angles DFE, FEC are equal to two right angles b; therefore, DEF, FEB are less than tworight angles d; therefore, FD, EB will meet one another, which let be

d. Cor. 176 in G. But CEF is a rightangle, and CEB half a rightangle; there. 1. fore, FEB is half a right angle, and EGFislikewise half a rightangle; therefore, EF is equal to IG, and the square of EG equal e 6.1. to the squares of EF, FG a, or double the square of EF, or CD; therefore, the squares of AE, EG are double the squares of AC, CD; but the square of AG is equal to the squares of AE, EG,

and

c Def, 1.

Book II' and likewise equal to the squares of AD, DG; for the angles nAEG, and ADG are each right ones; therefore, the squares of

AD, DG, or DB its equal, are double the squares of AD, CD.
Wherefore, &c.

PRO P. XI. PRO B.

T'

O cut a given right line so, that the rectangle contained un

der the whole line, and one of the parts, be equal to the Square of the other part.

A 40. I.

b 6.
c Const.

Upon any given right line, as AB, describe the square ABDC ~; bifečt AC in E;join EB, and produce EA to F; make EF equal to EB; upon AF, describe the square FGHA, and produce GH to K; then AB is so cut in the point H, that the rectangle under AB, BH is equal to the square of AH.

For the rectangle under CF, FA, together with the square of AE, is equal to the square of EFb; but EF is equal to EBC, and the square of EB is equal to the squares of BA, AE d ; therefore, the rectangle under CF, FA, together with the square of AE, are equal to the squares of BA, AE. Take the square of AE from both, there remains the rectangle under CF, FA, that is, the rectangle under CF, FG, that is, FK, equal to the square of AD. Take AK from both, there remains FH equal to HD ; but FH is the square of AH, and HD the rectangle under AB, BH, for BD is equal to AB. Wherefore, &c.

d 47. 1.

PRO P. XII. THEO R.

[ocr errors]

Nevery obtuse angled triangle, the square of the side subtending

the obtufe angle, is greater than the squares of the sides con

taining the obtuse angle, by twice a rectangle under one of the fides containing the obtuse angle, and i hat part of the side produced, lying betwixt the obtufe angle, and perpendicular let fall from the opposite angle.

a 12. I,

Let BAC be the obtufe angle of the triangle ABC; produce the fide CA till it meet the peopendicular BD, let fall from the point B 2. The square of BC is greater than the squares of BA, AC, by twice the rectangle under CA, AD.

For the square of BC is equal to the squares of BD, DC); but the square of DC is equal to the squares of DA, AC, and

[ocr errors]

twice the rectangle under AD, AC; but the square of AB Book II. is equal to the squares of BI), DAb; therefore the square of BC is equal to the squares of BA, AC, and twice the rectangle 4.

b 47. i. under DA, AC; therefore the square of BC is greater than the squares of BA, AC, by twice the rectangle under DA, AC. Wherefore, &c.

PRO P. XIII. THE O R.

I Nevery acute angled triangle, the square of the fide subtending

the acute angle, is less than the squares of the fide containing the acute angle, by twice a rectangle contained under one of the fides about the acute angle, and that part of the side lying between the acute angle and the perpendicular let fall from the opposite angle.

Let B be an acute angle in the triangle ABC ; from the angle A let fall the perpendicular AD, cutting BC in D; a 12. I. the square of AC is less than the squares of AB, BC, by twice the rectangle under CB, BD.

For the square of AC is equal to the squares of AD, DC6; 47. 1: and the square of AB is equal to the squares of AD, DB b; but the squares of BC, BD, are equal to twice the rectangle under BC, BD, together with the square of DC“; therefore the c . squares of AB, BC, are equal to the squares of AD, DC, and twice the rectangle under CB, BD, but the square of AC is equal to the squares of AD, DC; therefore the square of AC is less than the squares of AB, BC, by twice the rectangle under CB,BD. Therefore, &c.

PRO P. XIV. P R O B,

T

O make a square equal to a given right lined figure.

Make the rectangle BCDE equal to a given right lined figure A"; If BE be equal to ED, then BCDE is a square ; a 45.8. and what was required is done. If not, produce BE to F; make EF equal to ED, and bisect BF in Gb; with the center b 10. 1. G, and distance GB, describe a semicircle BHF; produce DE to H, and join GH.

Then
E

« ΠροηγούμενηΣυνέχεια »