*11. 1.

* 3. 1.

*31. 1.

*31. 1.

PROP. I. THEOR.

If there be two straight lines, one of which is divided 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 contained 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.

G

B

From the point B draw*
BF at right angles to BC, F
and make BG equal to A;

DEC

KLI

A

and through G draw* GH parallel to BC; and through D, E, C draw* DK, EL, CH parallel to BG. Then the rectangle BH is equal to the rectangles BK, DL, EH: but BH is contained by A, BC, for it is contained by GB, BC, and + Const. 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, *34. 1. DE, because DK, that is * BG, is equal to 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 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 rectangle con

tained by AB, BC, together with

the rectangle AB, AC, shall be A

C B

equal to the square of AB.

Upon AB describe the square

*46. 1.

*31. 1.

ADEB, and through C draw* CF,
parallel to AD or BE. Then AE
is equal to the rectangles AF, D
CE: but AE is the square of

AC;

FE

AB; and AF is the rectangle contained by BA, for it is contained by DA, AC, of which AD is equal to AB; and CE is contained by †30 Def. 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 f 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 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 A C

B

*46. 1.

square CDEB, and produce ED to F, and through A

AE is equal to the rectangles
AD, CE: but AE is the

*31. 1.

rectangle contained by AB, BC, for it is con

tained by AB, BE, of which BE is equal † to †30 Def.

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

1

*46. 1.'

BC; and AD is contained 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 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 *31. 1. join BD, and through C draw CGF parallel to AD or BE, and through G draw HK parallel † Const. to AB or DE. And because CF is parallel† to AD, and BD falls upon them, the exterior angle *29. 1. BGC is equal to the interior and opposite angle ADB; but ADB is equal to the angle ABD, †30 Def. because BA is equal to AD†, being sides of a square; wherefore the angle

*5.1.

*

*

+1 Ax. CGB is equal to the angle A
CBG; and therefore the side
BC is equal to the side CG: H
but CB is equal also to GK, and
CG to BK; wherefore the figure
CGKB ist equilateral: it is D'
likewise rectangular; for, since

*6. 1.

f1 Ax.

CG is parallel to BK, and CB meets them, +29. 1. therefore the angles KBC, GCB are equal † to †30 Def. two right angles: but KBC is a ↑ right angle; +3 Ax. wherefore GCB is at right angle: and there*34. 1. fore also the angles *, CGK, GKB opposite to 1. Ax. these, are right angles; and therefore CGKB is rectangular but it is also equilateral, as was +30Def. demonstrated; wherefore it is at square, and it is upon the side CB: for the same reason HF also

and

:

is a square, and it is upon the side HG, which is equal to AC therefore HF, CK are the +34. 1. squares of AC, CB: and because the complement AG is equal * to the complement GE, and that *43. 1. AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also equal + +30Def. to the rectangle AC, CB; wherefore AG, GE are +1ax. 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 the up 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 parallelograms about the diameter of a square are likewise squares.

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, shall be equal to the square of CB.

Ax.

Upon CB describe* the square CEFB, join *46. 1. BE, and through D draw* DHG parallel to CE *31. 1. 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 equal to the complement HF, *43. 1. to each of these add DM; therefore the whole CM is equal to the whole DF: but CM is +2 Ax.

*36. 1. equal to AL, because AC is equal† to CB; + Hyp. 1 Ax. therefore also AL is equal† to DF: to each of these add CH, and the

+2 Ax. whole AH is equal to

DF and CH: but AH is the rectangle contained by *Cor. 4. AD, DB, for DH is equal* 2. and 30 to DB; and DF together Def. with CH is the gnomon

2. and 34. 1.

+2 Ax.

*

C DB

L

H

M

K

E

G F

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

46. 1.

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.

PROP. 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.

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, join *31.1. DE, and through B draw* BHG parallel to CE