and AF is the rectangle contained by BA, AC; for it is con- Book II. 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. Q. E. D.

a

a 46. I.

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 forefaid part.

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 describe a the square

B

А С CDĖB, 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; 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 contained by AC, CB, for CD is equal to CB; and DB is the F D

E square of BC; therefore the rectangle AB, BC is equal to the rectangle AC, CB together with the square of BC. If therefore a ftraight line, &c. Q. E. D.

b 31,1.

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 fquares of AC, CB and to twice the rectangle contained by AC, CB..

Upon

D 3

e

f 34. 1.

Upon AB describe " the square ADEB, and join BD, and thro' C drawb CGF parallel to AD or BE, and thro' 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 d to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB

A С B
is equal to the angle GBC; and there-
fore the fide BC is equal to the
fide CG: But CB is equal falso to

G
GK, and CG to BK; wherefore H

K
the figure CGKB is equilateral : It is
likewise rectangular; for CG is pa-
rallel to BK, and CB meets them ;
the angles KBC, GCB are therefore
equal to two right angles; and KBC D F E
is a right angle; wherefore GCB is a right angle; and therefore
also the angles f CGK, GKB opposite to these are right angles,
and CGKĎ 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
fide HG which is equal to AC: Therefore HF, CK are the
squares of AC, CB; and becaufe the complement AG is equal
s to the complement GE, and that AG is the rectangle contain-
ed 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. R E. D.

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

PROP.

с

Book 11. PROP. V. THE O R. 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.

Upon CB describe a the square CEFB, join BE, and thro' a 46. 1. D draw 5 DHG parallel to CE or BF ; and thro' H draw b 31 1. KLM parallel to CB or EF; and also thro' 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; C 43. I therefore the whole CM is equal to the whole DF; but CM is equal d to AL, А С D B

d 36. T because AC is equal to

L H CB; therefore allo AL is K

M 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

E G F AD, DB, for DH is equal

to DB; and DF together with CH is the gnomon CMG; Cor. 4, 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. Q. É. D.

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

AN

с

Book II.

PRO P. VI.

THEOR.

а

a 46. r.

с

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 of 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 bifected 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 CD describe the square CEFD, join DE, and thro' b 31.1. B draw b BHG parallel to CE or DF, and thro' H draw KLM

parallel to AD or EF, and also thro' A draw AK parallel to CL
or DM: And because AC
is cqual to CB, the rectangle

A C B D C 36. 1.

AL is equal to CH; but d 43. I. CH is equald to HF; there

L HI fore alfa Al is equal to K

M HF: 'To each of these add CM; therefore the whole AM is equal to the gnomon CMG: And AM is the rectangle contained by

E G F AD, DB, for DM is equal € Cor. 4. 20 C to DB : Therefore the gnomon CMG is equal to the reçt

angle 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.

PRO P. VII. TH LO R.
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

the point C; the squares of AB, BC are equal to twice the Book II, rectangle AB, BC together with the square of AC.

Upon AB describe a the square. ADEB, and construct the a 46. I. figure as in the preceeding propositions: And because AG is equal to GE, add to each of them CK ; the whole AK is b 43. 1. therefore cqual to the whole CE; therefore. AK, CE are double of A С AK: But AK, CE are the gnomon

B AKF together with the square CK; therefore the gnomon AKF, toge-H

G

K ther with the square CK, is double of AK: But twice the rectangle AB, BC is double of AK, for BK is e

c Cor. 4.2. qual to BC: Therefore the gnomon AKF, together with the square

D

E • CK, is equal to twice the rectangle

F AB, BC: To each of these equals add HF, which is equal to the square of AC; therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC and the square of AC : But the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the Tquares 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 Itraight line, &c. Q. E. D.

IF straight line be divided into any two parts, four

times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made

up

of the whole and that part.

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

Produce AB to D, fo that BD be equal to CB, and upon AD describe the square AEFD; and construct two figures such as in the preceeding. Because C} is equal to BD, and that CB is equal to GK, and BD to KN; therefore GK is a 31.1.