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 and produced to the point D:

bisected at the point C, the rectangle AD, DB,

together with the square on CB, shall be equal to the

square on CD.

EF; and through A draw

AK parallel to CL or DM.

[I. 31.

Then, because AC is equal to CB,

[Hypothesis.

the rectangle AL is equal to the rectangle CH;

[I. 36.

but CH is equal to HF;

[I. 43.

therefore also AL is equal to HF.

[Axiom 1.

To each of these add CM;

therefore the whole AM is equal to the gnomon CMG. [Ax. 2. But AM is the rectangle contained by AD, DB, for DM is equal to DB. [II. 4, Corollary. Therefore the rectangle AD, DB is equal to the gnomon CMG. [Axiom 1. To each of these add LG, which is equal to the square on CB. [II. 4, Corollary, and I. 34. 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.

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

On AB describe the square ADEB, and construct the figure as in the preceding propositions.

Then AG is equal to GE; [I. 43. to each of these add CK; therefore the whole AK is equal to the whole CE;

therefore AK, CE are double of AK.

H

D

K

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, for BK is equal to BC. [II. 4, Corollary. 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 on AC. [II. 4, Corollary, and I. 34. together with the squares

Therefore the gnomon AKF, CK, HF, is equal to twice the rectangle AB, BC, together with 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, 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 at the point C: 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.

and that CB is equal to GK, and BD to KN, therefore GK is equal to KN.

For the same reason PR is equal to RO.

[I. 34.

[Axiom 1.

And because CB is equal to BD, and GK to KN, the rectangle CK is equal to the rectangle BN, and the rectangle GR to the rectangle RN.

[I. 36.

But CK is equal to RN, because they are the complements of the parallelogram CO;

therefore also BN is equal to GR.

[I. 43.

[Axiom 1.

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

that is to GP;

therefore CG is equal to GP.

[II. 4, Corollary.

[Axiom 1.

And because CG is equal to GP, and PR to RO, the rectangle AG is equal to the rectangle MP, and the rectangle PL to the rectangle RF.

[I. 36.

But MP is equal to PL, because they are the complements of the parallelogram ML;

therefore also AG is equal to RF.

[I. 43.

[Axiom 1.

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

And it was shewn that the four CK, BN, GR and RN are quadruple of CK; therefore the eight rectangles which make up 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 shewn to be quadruple of AK.

Therefore four times the rectangle AB, BC is equal to the gnomon AOH.

[Axiom 1. To each of these add XH, which is equal to the square on AC. [II. 4, Corollary, and I. 34. 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 the square 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 to the square on the line made of AB and BC together.

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

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 at the point C, and into two unequal parts at the point D: the squares on AD, DB shall be together double of the squares on AC, CD.

From the point C draw CE at right angles to AB, [I. 11. and make it equal to AC or CB, [I. 3. and join EA, EB; through D draw DF parallel to CE, and through F draw FG parallel to BA;

and join AF.

C

[I. 31.

Then, because AC is equal to CE,

the angle EAC is equal to the angle AEC.

[Construction.

[I. 5.

And because the angle ACE is a right angle, [Construction.

the two other angles AEC, EAC are together equal to one right angle;

and they are equal to one another;

therefore each of them is half a right angle.

[I. 32.

For the same reason each of the angles CEB, EBC is half a right angle.

Therefore the whole angle AEB is a right angle.

And because the angle GEF is half a right angle, and the angle EGF a right angle, for it is equal to the interior and opposite angle ECB;

[I. 29.

therefore the remaining angle EFG is half a right angle. Therefore the angle GEF is equal to the angle EFG, and the side EG is equal to the side GF. Again, because the angle at B is half a right angle, and the

[I. 6.