On BC describe the square CDEB; produce ED to F, and through AA draw AF parallel to CD or BE. [I. 31.

Then the rectangle AE is equal to the rectangles AD, CE. But AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC;

ལ

and AD is contained by AC, CB, for CD is equal to CB; and CE is the square on BC.

Therefore the rectangle AB, BC is equal to the rectangle AC, CB, together with the square on BC.

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

PROPOSITION 4. THEOREM.

If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the two parts.

Let the straight line AB be divided into any two parts at the point C: the square on AB shall be equal to the squares on AC, CB, together with twice the rectangle contained by AC, CB.

On AB describe the square ADEB; [I. 46. join BD; through C draw CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. [1.31.

Then, because CF is parallel to AD, and BD falls on them, the exterior angle CGB is equal to the interior and opposite angle ADB ;

[I. 29.

H

[I. 5.

but the angle ADB is equal to the angle ABD, because BA is equal to AD, being sides of a square; therefore the angle CGB is equal to the angle CBG; [Ax. 1. and therefore the side CG is equal to the side CB. But CB is also equal to GK, and CG to BK ; therefore the figure CGKB is equilateral.

[I. 6. [I. 34.

It is likewise rectangular. For since CG is parallel to BK, and CB meets them, the angles KBC, GCB are together equal to two right angles.

But KBC is a right angle.

Therefore GCB is a right angle.

[I. 29.

[I. Definition 30.

[Axiom 3.

And therefore also the angles CGK, GKB opposite to these are right angles.

Therefore CGKB is rectangular;
and it has been shewn to be equi-
lateral; therefore it is a square, and
it is on the side CB.

For the same reason HF is also a
square,
and it is on the side HG,
which is equal to AC.

[I. 34. and Axiom 1.

[I. 34.

D

B

And because the complement AG is equal to the_complement GE;

[I. 43. and that AG is the rectangle contained by AC, CB, for CG is equal to CB;

therefore GE is also equal to the rectangle AC, CB. [Ax. 1. Therefore AG, GE are equal to twice the rectangle AC, CB. And HF, CK are the squares on AC, CB.

Therefore the four figures HF, CK, AG, GE are equal to the squares on AC, CB, together with 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, together with twice the rectangle AC, ĈB.

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

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

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 at the point C, and into two unequal parts at the point D: the rectangle AD, DB, together with the square on CD, shall be equal to the square on CB.

[I. 46.

On CB describe the square CEFB; join BE; through D draw DHG parallel to CE or BF; through Hdraw KLM parallel to CB or EF; and through A draw AK parallel to CL or BM.

[I. 31.

Then the complement CH is equal to the complement

HF;

[I. 43.

to each of these add DM; therefore the whole CM is equal to the whole DF.

But CM is equal to AL,

because AC is equal to CB.

Therefore also AL is equal to DF.

[Axiom 2.

[I. 36. [Hypothesis.

[Axiom 1.

To each of these add CH; therefore the whole AH is equal to DF and CH. [Axiom 2. But AH is the rectangle contained by AD, DB, for DH is equal to ᎠᏴ ; [II. 4, Corollary. 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 to the square on CD. [II. 4, Corollary, and I. 34. Therefore the gnomon CMG, together with LG, is equal to the rectangle AD,DB, together with the square on CD. [Ax.2. 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.

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

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 at the point C, and produced to the point D: the rectangle AD, DB, together with the square on CB, shall be equal to the square on CD.

B

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.

the

If a straight line be divided into any two parts, 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

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 ÃB, 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.