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Let A and BC be two straight lines; and let BC be divided into any parts BD, DE, EC, in the points D, E.

Then the rectangle contained by the straight lines A and BC shall be equal to the rectangle contained by A and BD, together with that contained by A and DE, and that contained by A and EC.

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Construction. 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 (I. 31), and through D, É, C draw DK, EL, CH parallel to BG, meeting GH in K, L, H.

Demonstration. Then the rectangle BH is equal to the rectangles BK, DL, EH ; and

1. BH is contained by A and BC,

for it is contained by GB, BC, and GB is equal to A; and the rectangle

2. BK is contained by A, BD,

for it is contained by GB, BD, of which GB is equal to A; also

3. DL is contained by A, DE,

because DK—that is, BG (I. 34)—is equal to A; and in like manner 4. The rectangle EH is contained by A, EC;

therefore the rectangle contained by A, BC, is equal to the several 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 in the point C.

E

Then the rectangle contained by AB, BC, together with that contained by AB, AC, shall be equal to the square on AB.

B

D

Construction. Upon AB describe the square ADEB (I. 46), and through C draw CF parallel to AD or BE (I. 31), meeting DE in F. Demonstration. Then AE is equal to the rectangles AF, CE. And AE is the square on AB, and

1. AF is the rectangle contained by BA, AC;

for it is contained by DA, AC; of which DA is equal to AB; and 2. CE is contained by AB, BC;

for BE is equal to AB; therefore

3. The rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square on AB.

Therefore, 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 in the point C.

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Then the rectangle AB, BC shall be equal to the rectangle AC, CB, together with the square on BC.

Construction. Upon BC describe the square CDEB (I. 46), and produce ED to F; through A draw AF parallel to CD or BE (I. 31), meeting EF in F

Demonstration. Then

And

1. The rectangle AE is equal to the rectangles AD, CE.

2. AE is the rectangle contained by AB, BC,

for it is contained by AB, BE, of which BE is equal to BC; and 3. AD is contained by AC, CB,

for CD is equal to CB (Def. 30), and CE is the square on BC; therefore

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

If therefore a straight line be divided, &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 parts.

Let the straight line AB be divided into any two parts in C.

Then the square on AB shall be equal to the squares on AC and CB, together with twice the rectangle contained by AC, CB.

[blocks in formation]

Construction. Upon AB describe the square ADEB (I. 46); join BD; through C draw CGF parallel to AD or BE (I. 31), meeting BD in G, and DE in K; and through G draw HGK parallel to AB or DE, meeting AD in H, and BE in K.

Demonstration. Then, because CF is parallel to AD, and BD falls upon them, therefore

1. The exterior angle BGC is equal to the interior and opposite angle BDA (I. 29),

but

H

K

2. The angle BDA is equal to the angle DBA (I. 5), because BA is equal to AD, being sides of a square; wherefore 3. The angle BGC is equal to the angle DBA or GBC;

and therefore

4. The side BC is equal to the side CG (I. 6),

but BC is equal also to GK, and CG to BK (I. 34); wherefore 5. The figure CGKB is equilateral.

It is likewise rectangular; for, since CG is parallel to BK, and BC meets them, therefore

6. The angles KBC, BCG are equal to two right angles (I. 29), but the angle KBC is a right angle (Def. 30, constr.), wherefore

7. BCG is a right angle;

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

8. The figure CGKB is rectangular.

But it is also equilateral, as was demonstrated, wherefore

9. CGKB is a square,

and it is upon the side CB. For the same reason,

10. HF is a square,

and it is upon the side HG, which is equal to AC (I. 34). Therefore 11. 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

12. AG is the rectangle contained by AC, CB;

for GC is equal to CB; therefore also

13. GE is equal to the rectangle AC, CB;

wherefore

14. AG, GE are equal to twice the rectangle AC, CB; and HF, CK are the squares on AC, CB; wherefore

15. 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

16. 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 5.-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.

[blocks in formation]

Construction. Upon CB describe the square CEFB (I. 46): join BE; through D draw DHG parallel to CE or BF (I. 31), meeting BE 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.

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