Upon CB describe the square CEFB, (1. 46.) join BE, through D draw DHG parallel to CE or BF, (1. 31.) meeting DE in H, and EF in G,

and through H draw KLM parallel to CB or EF, meeting CE in L, and BF in M;

also through A draw AK parallel to CL or BM, meeting MLK in K. Then because the complement CH is equal to the complement HF, (1. 43.) to each of these equals add DM;

therefore the whole CM is equal to the whole DF;

but because the line AC is equal to CB,
therefore AL is equal to CM, (1. 36.)
therefore also AL is equal to DF;

to each of these equals add CH,

and therefore 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 equals add LG, which is equal to the square on CD; (II. 4. Cor.)

therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square on CD:

but the gnomon CMG and LG make up the whole figure CEFB, which is the square on CB;

therefore the rectangle AD, DB, together with the square on CD is equal to the square on CB.

Wherefore, if a straight line, &c.

Q. E.D.

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

PROPOSITION VI. THEOREM.

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 on half the line bisected, is equal to the square on 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. Then the rectangle AD, DB, together with the square on ĈB, shall be equal to the square on CD.

Upon CD describe the square CEFD, (1. 46.) and join DE, through B draw BHG parallel to CE or DF, (1. 31.) meeting DE in H, and EF in G;

through H draw KLM parallel to AD or EF, meeting DF ir. M, and CE in L;

and through A draw AK parallel to CL or DM, meeting MLK in K.

Then because the line AC is equal to CB,

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

to each of these equals add CM;

therefore the whole AM is equal to the gnomon CMG:
but AM is the rectangle contained by AD, DB,
for DM is equal to DB: (II. 4. Cor.)

therefore the gnomon CMG is equal to the rectangle AD, DB: to each of these equals add LG which is equal to the square on CB; therefore the rectangle AD, DB, together with the square on 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 on CD;

therefore the rectangle AD, DB, together with the square on CB, is equal to the square on CD.

Wherefore, if a straight line, &c.

Q. E. D.

PROPOSITION VII. THEOREM.

If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part.

Let the straight line AB be divided into any two parts in the point C. Then the squares on AB, BC shall be equal to twice the rectangle AB, BC, together with the square on AC.

Upon AB describe the square ADEB, (1. 46.) and join BD; through Cdraw CF parallel to AD or BE (1. 31.) meeting BD in G, and DE in F

through G draw HGK parallel to AB or DE, meeting AD in H, and BE in K.

Then because AG is equal to GE, (1. 43.)
add to each of them CK;

therefore the whole AK is equal to the whole CE;
and therefore AK, CE, are double of AK:

but AK, CE, are the gnomon AKF and the square CK; therefore the gnomon AKF and the square CK are double of AK: but twice the rectangle AB, BC, is double of AK, for BK is equal to BC; (II. 4. Cor.)

therefore the gnomon AKF and the square CK, are equal to twice the rectangle AB, BC;

to each of these equals add HF, which is equal to the square on AC, therefore the gnomon AKF, and the squares CK, HF, are equal to twice the rectangle AB, BC, and the square on AC;

but the gnomon AKF, together with the squares CK, HF, make

up the whole figure ADEB and CK, which are the squares on AB and BC;

therefore the squares on AB and B C'are equal to twice the rectangle

AB, BC, together with the square on AC.

Wherefore, if a straight line, &c. Q. E. D.

If a 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 on the other part, is equal to the square on 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. Then four times the rectangle AB, BC, together with the square on AC, shall be equal to the square on the straight line made up of AB and BC together

Produce AB to D, so that BD be equal to CB, (1. 3.) upon AD describe the square AEFD, (1. 46.) and join DE, through B, C, draw BL, CH parallel to AE or DF, and cutting DE in the points K, P respectively, and meeting EF in L, H; through K, P, draw MGKN, XPRO parallel to AD or EF. Then because CB is equal to BD, CB to GK, and BD to KN; therefore GK is equal to KN;

for the same reason, PR is equal to RO;

and because CB is equal to BD, and GK to KN, therefore the rectangle CK is equal to BN, and GR to KN; (1. 36.) but CK is equal to RN, (1. 43.)

because they are the complements of the parallelogram CO;
therefore also BN is equal to GR;

and the four rectangles BN, CK, GR, RN, are equal to one another, and so are quadruple of one of them CK.

Again, because CB is equal to BD, and BD to BK, that is, to CG; and because CB is equal to GK, that is, to GP;

therefore CG is equal to GP.

And because CG is equal to GP, and PR to RO, therefore the rectangle AG is equal to MP, and PL to RF; but the rectangle MP is equal to PL, (1. 43.)

because they are the complements of the parallelogram ML: wherefore also AG is equal to RF:

therefore the four rectangles AG, MP, PL, RF, are equal to one another, and so are quadruple of one of them AG.

And it was demonstrated, that the four CK, BN, GR, and RN, aro quadruple of CK:

therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK.

And because AK is the rectangle contained by AB, BC,

for BK is equal to BC;

therefore four times the rectangle AB, BC is quadruple of AK: but the gnomon AOH was demonstrated to be quadruple of AK; therefore four times the rectangle AB, BC is equal to the gnomon AOII; to each of these equals add XH, which is equal to the square on AC; therefore four times the rectangle AB, BC, together with the square on AC, is equal to the gnomon AOH and the square XH; but the gnomon AOH and XH make up the figure AEFD, which is the square on AD;

therefore four times the rectangle AB, BC together with the square on AC, is equal to the square on AD, that is, on AB and BC added together in one straight line.

Wherefore, if a straight line, &c, Q.E.D.

PROPOSITION IX. THEOREM.

If a straight line be divided into two equal, and also into two unequal parts; the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of section.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D.

Then the squares on AD, DB together, shall be double of the squares on AC, CD.

From the point C draw CE at right angles to AB, (1. 11.) make CE equal to AC or CB, (1. 3.) and join EA, EB; through D draw DF parallel to CE, meeting EB in F, (1. 31.) through Fdraw FG parallel to BA, and join AF. Then, because AC is equal to CĚ,

therefore the angle AEC is equal to the angle EAC; (1. 5.)
and because ACE is a right angle,

therefore the two other angles AEC, EAC of the triangle are together equal to a right angle; (1. 32.)

and since they are equal to one another;

therefore each of them is half a right angle.

For the same reason, each of the angles CEB, EBC is half a right angle; and therefore the whole AEB is a right angle.

And because the angle GEF is half a right angle,
and EGF a right angle,

for it is equal to the interior and opposite angle ECB, (1. 29.)
therefore the remaining angle EFG is half a right angle;
wherefore the angle GEF is equal to the angle EFG,
and the side GF equal to the side EG. (1. 6.)

Again, because the angle at B is half a right angle,
and FDB a right angle,

for it is equal to the interior and opposite angle ECB, (1. 29.)
therefore the remaining angle BFD is half a right angle;
wherefore the angle at B is equal to the angle BFĎ,
and the side DF equal to the side DB. (1. 6.)
And because AC is equal to CE,

the square on AC is equal to the square on CE;

therefore the squares on AC, CE are double of the square on A C; but the square on AE is equal to the squares on AC, CE, (1. 47.) because ACE is a right angle;

therefore the square on AE is double of the square on AC.
Again, because EG is equal to GF,

the square on EG is equal to the square on GF;

therefore the squares on EG, GF are double of the square on GF; but the square on EF is equal to the squares on EG, GF; (1. 47.) therefore the square on EF is double of the square on GF; and GF is equal to CD; (I. 34.)

therefore the square on EF is double of the square on CD; but the square on AE is double of the square on AC; therefore the squares on AE, EF are double of the squares on AC, CD; but the square on AF is equal to the squares on AE, EF, because AEF is a right angle: (1.47.)

therefore the square on AF'is double of the squares on AC, CD: but the squares on AD, DF are equal to the square on AF; because the angle ADF is a right angle; (1. 47.)

therefore the squares on AD, DF are double of the squares on 4 C, CD; and DF is equal to DB;

therefore the squares on AD, DB are double of the squares on AC, CD. If therefore a straight line be divided, &c.

PROPOSITION X. THEOREM.

Q.E. D.

If a straight line be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D.

Then the squares on AD, DB, shall be double of the squares on AC, CD.

From the point C draw CE at right angles to AB, (1. 11.)
make CE equal to AC or CB, (1. 3.) and join AE, EB;
through E draw EF parallel to AB, (1. 31.)

and through D draw DF parallel to CE, meeting EF in F.