wherefore the angle BGC is equal to the angle DBA or GBC; and therefore the side BC is equal to the side CG; (1. 6.) but BC is equal also to GK, and CG to BK; (1. 34.) wherefore the figure CGKB is equilateral.

It is likewise rectangular;

for, since CG is parallel to BK, and BC meets them, therefore the angles KBC, BCG are equal to two right angles; (I. 29.) but the angle KBC is a right angle; (def. 30. constr.) wherefore BCG is a right angle:

and therefore also the angles CGK, GKB, opposite to these, are right angles; (1. 34.)

wherefore 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 is a square,

and it is upon the side HG, which is equal to AC. (1.34.) Therefore the figures HF, CK, are the squares on AC, CB. And because the complement AG is equal to the complement GE, (I. 43.)

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 on AC, CB;

wherefore the four figures HF, CK, AG, GE, are equal to the squares on AC, CB, and twice the rectangle AC, CB:

but HF, CK, AG, GE make up the whole figure ADEB, which is the square on AB;

therefore the square on AB is equal to the squares on AC, CB, and twice the rectangle AC, CB.

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

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

PROPOSITION V. THEOREM.

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 on the line between the points of section, is equal to the square on half the line.

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 rectangle AD, DB, together with the square on CD, shall be equal to the square on CB.

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 4 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 AĎ, 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 CB, 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, (I. 31.) meeting DE in H, and EF in G;

through H draw KLM parallel to AD or EF, meeting DF in 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 CH, (1. 36.) but CH is equal to HF; (1. 43.) therefore AL is equal to HF;

to each of these equals add CM;

CMG:

therefore the whole AM is equal to the gnomon
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 C draw 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 BC' 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 RN; (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, are 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 AOH; 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, (I. 31.) through Fdraw FG parallel to BA, and join AF. Then, because AC is equal to CE,

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 ÈFG 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.)