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.

*

D

KLH

*11. 1.

3.1.

31. 1.

31. 1.

From the point B draw BF at right angles to BC, and make BG equal to 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 GB is equal to A; and BK is contained by A, BD, for it is +Constr. contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, because DK, that is * BG, *34.1. 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.

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 contained by AB, BC, together with the rectangle AB, AC shall be equal to the square of AB.

C B

* 46.1.

*31. 1.

Upon AB describe the square ADEB, and 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 contained by DA, AC, of which AD is equal † to AB: and CE is contained † 30 Def. by AB, BC, for BE is equal to AB: therefore the rectangle AC, contained by AB, AC, together with the rect

D

FE

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.

46. 1.

* 31. 1.

angle 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 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.

B

E

Upon BC describe the square CDEB, and produce ED to F, and through A 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 AB, BE, of which BE is +30 Def. equal + to 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 á straight line, &c. Q. E. D.

*46. 1.

31. 1.

+ Constr.

* 29.1.

*5.1.

+ 30 Def.

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 join BD, and through C draw * CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And because CF is parallel + to AD, and BD falls upon them, the exterior angle BGC is equal to the interior and opposite angle ADB; but ADB is equal to the angle ABD, because BA is equal to AD†, being sides of

*

*

a

*

C B†1 Ax. *6.1.

D FE

K

*34. 1.

†1 Ax.

square; wherefore the angle CGB is equal + to the angle CBG; and therefore the side BC is equal to the side CG: but CB is equal also to GK, and CG H to BK; wherefore the figure 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 + 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 at square, and it is upon the side CB: for the same +30 Def. reason HF also 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

*

29. 1. 30 Def. 34. 1. † 3 Ax. and 1 Ax.

is equal to the complement GE, and that AG is the * 43. 1. rectangle contained by AC, CB, for GC is equal to + 30 Def. CB; therefore GE is also equal+ to the rectangle AC, †1 Ax. 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, † 1 Ax. 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.

*

Upon CB describe the square CEFB, join BE, and * 46. 1. through D draw* DHG parallel to CE or BF; and 31.1. through H draw KLM parallel to CB or EF; and also

* 43. 1. +2 Ax.

36. 1.

† Hyp. † 1 Ax.

† 2 Ax.

*Cor. 4. 2.
& 30 Def.

† 1 Ax.
*Cor. 4. 2.
and 34. 1.

† 2 Ax:

* 46. 1.

* 31. 1.

+ Hyp.

36.1.

$43.1%

† 1 As.

*

*

C DB

LH

M

E G F

through A draw AK parallel to CL or BM. And be-
cause the complement CH is equal to the complement
HF, to each of these add DM; therefore the whole CM
is equal to the whole DF: but CM is equal to AL,
because AC is equal to CB; therefore also AL is equal+
to DF: to each of these add CH,
and the whole AH is equal+ to DF
and CH: but AH is the rectangle
contained by AD, DB, for DH is
equal* 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, toge-
ther with LG is equal to the rectangle AD, DB toge-
ther 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, toge-
ther with the square of CD, is equal to the square of
CB. Wherefore, if a straight line, &c. Q. E. D.

*

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 DE, and through B draw BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK parallel to CL or DM. And because AC is equal to CB, the rectangle AL is equal to CH; but CH is K equal to HF; therefore also AL is equal to HF: to each of these

*

*

add CM; therefore the whole AM

B D

H

M

E

G F

*

† 1. Ax,

& 30 Def. + Cor. 4.2. and 34. 1.

is equal to the gnomon CMG: but AM is the rect- † 2 Ax. angle contained by AD, DB, for DM is equal to DB: * Cor. 4.2. therefore the gnomon CMG is equal to the 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.

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 the point C: the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square of AC.

*

*

A

H

C

+2 Ax.

43. 1.

+ 1 Ax.

Upon AB describe the square ADEB, and con- * 46.1. struct the figure as in the preceding propositions. And because AG is equal to GE, add to each of them CK; therefore the whole AK is equal to the whole CE; therefore AK, CE, are double of AK: but AK, CE, are the gnomon AKF, together with the square CK; therefore the gnomon AKF, together with the square CK, is + double of AK: but twice the rectangle AB, BC is double of AK, D for BK is equal to BC; therefore the gnomon AKF, together with the square CK, is equal † to twice the rectangle AB, BC: to each of these equals add HF, which is equal + to the square of AC: there- † Cor. 4. 2. fore the gnomon AKF, together with the squares CK, & 34. 1. HF, is equal to twice the rectangle, AB, BC, and the † 2 Ax. square of AC: but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC; therefore the squares of AB and BC are equal to twice the rectangle AB, BC, together with the square of AC. Wherefore if a straight line, &c. Q. E. D.

* Cor. 4.2. & 30 Def. + 1 Ax.