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 by the whole and one of the parts is equal with the rectangle contained by the two parts, together with the square of the aforesaid part.

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

с

Upon BC describe (1.46.) the square A CDEB, and produce ED to F, and through a draw (I. 31.) AF parallel to CD or BE; then the rectangle AE is equal to the rectangles AD, CE; and AE is the rectangle contained F D

B

E

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

PROP. IV. THEOR.

If

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, BC, together with twice the rectangle con-` tained by AC, CB.

Upon AB describe (1. 46.) the square ADEB, and join BD, and through C draw (I. 31.) 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 (1. 29.) to the interior and opposite angle ADB; but ADB is equal (1. 5.) to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle GBC; and therefore the side BC is equal (1. 6.) to the side D CG: But CB is equal (1. 34.) also to GK, and CG to BK; wherefore the figure CGKB is equilateral: It H is likewise rectangular; for CG is parallel to BK, and CB meets them; the angles KBC, GCB are there

G

F

E

K

fore equal to two right angles; and KBC is a right angle; wherefore GCB is a right angle; and therefore also the angles (1. 34.) 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 (1. 43.) to the complement GE, and that AG is the rectangle 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 ADE B, 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.

If a straight line bé 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.

K

C

D B

H

M

Upon CB describe (1. 46.) the square CEFB, join BE, and through D draw (1. 31.) 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 equal (1. 43.) to the complement HF to each of these add DM; therefore the whole CM A is equal to the whole DF; but CM is equal (1.36.) 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, BD, for DH is equal (II. 4. Cor.) 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

E

G F

(11. 4. Cor.) to the square of CD, therefore the gnomon GMG, 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. 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 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, is equal to the square of CD.

Upon CD describe (1. 46.) the square CEFD, join DE, and through в draw (I. 31.) 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: cause AC is equal to

and be

A

C

BD

CB, the K

L

H

M

rectangle AL is equal (1. 36.) to CH; but CH is equal (1. 43.) to HF; therefore also AL is equal

E

G F

to HF: To each of these add CM; therefore the whole AM is equal to the gnomon CMG: And AM is the rectangle contained by AD, DB, for DM is

equal (II. 4. Cor.) to DB: 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.

If a straight line be divided into any two parts, the squares of the whole line, and 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 are equal to twice the rectangle AB, BC, together with the square of AC.

A

C

B

G

K

Upon AB describe (1. 46.) the square ADEB, and construct the figure as in the preceding propositions: And because AG is equal (1 43.) to GE, add to each of them CK; the whole AK is therefore equal to the whole CE; therefore AK, CE are double of AK; but AK, CE are the H 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, for BK is equal (I. 4. Cor.) to BC: Therefore the gnomon AKF, together with the square CK, is equal to twice the

D

F

E