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Thus the parallelogram HG, A together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon.

H

K

B

PROPOSITION 1. THEOREM.

If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Let A and BC be two straight lines; and let BC be divided into any number of parts at the points D, E: the rectangle contained by the straight lines A, BC, shall be equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC.

From the point B draw BF at right angles to BC; [I. 11. and make BG equal to A; [I. 3. through G draw GH parallel to BC; and through D, E, C draw DK, EL, CH, parallel to BG.

[I. 31.

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B

K

A

But BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A.

[Construction.

And BK is contained by A, BD, for it is contained by GB, BD, and GB is equal to A;

and DL is contained by A, DE, because DK is equal to BG, which is equal to ▲ ;

[I. 34.

and in like manner EH is contained by A, EC. Therefore the rectangle contained by A, BC is equal to the rectangles contained by A, BD, and by A, DE, and by A, EC.

Wherefore, if there be two straight lines &c. Q.E.D.

PROPOSITION 2. THEOREM.

If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square on the whole line.

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

[Note. To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC.]

On AB describe the square ADEB; [I. 46. and through C draw CF parallel to AD or BE.

[1.31.

Then AE is equal to the rectangles AF, CE. But AE is the square on AB.

And AF is the rectangle contained by BA, AC, for it is contained by DA, AC, of which DA is equal to BA;

and CE is contained by AB, BC, for BE is equal to AB. Therefore the rectangle AB, AC, together with the rectangle AB, BC, is equal to the square on AB. Wherefore, if a straight line &c. Q.E.D.

PROPOSITION 3. THEOREM.

If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square on the aforesaid part.

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

On BC describe the square CDEB; produce ED to F, and through AA draw AFparallel to CD or BE. [1.31.

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;

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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 ADEB;

square
[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.

K

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 ; but the angle ADB is equal to the angle ABD,

[I. 29.

[I. 5.

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 equilateral; 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.

Therefore HF, CK are the squares on AC, CB.

[I. 34. and Axiom 1.

A

H

B

K

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, CB.

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.

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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.

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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 DB; [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.

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