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6. 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

7. 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.

PROPOSITION 8.-Theorem.

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.

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Construction. Produce AB to D, so that BD be equal to CB (I. 3); upon AD describe the square AEFD (I. 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.

Demonstration. Then because CB is equal to BD, CB to GK, and BD to KN; therefore

1. GK is equal to KN; for the same reason,

2. PR is equal to RO; and because CB is equal to BD, and GK to KN, therefore

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E

H

F 3. The rectangle CK is equal to BN, and GR to RN (I. 36); but CK is equal to RN (I. 43), because they are the complements of the parallelogram CO; therefore also

4. BN is equal to GR; and therefore

5. 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

6. CG is equal to GP. And because CG is equal to GP, and PR to RO, therefore

7. The rectangle AG is equal to MP, and PL to RF; but the rectangle MP is equal to PL (I. 43), because they are the complements of the parallelogram ML; wherefore also

8. AG is equal to RF; therefore

9. 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

10. The eight rectangles which contain the gnomon AOH, are

quadruple of XK. And because AK is the rectangle contained by AB, BC, for BK is equal to BC; therefore

11. Four times the rectangle AB, BC is quadruple of AK;

but the gnomon AOH was demonstrated to be quadruple of AK; therefore

12. Four times the rectangle AB, BC is equal to the gnomon

Аон; to each of these equals add XH, which is equal to the square on AC; therefore

13. 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

14. 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 9.-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 an AC, CD.

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Construction. From the point C draw CE at right angles to AB (I. 11); make CE equal to AC or CB (I. 3), and join EA, EB; through D draw DF parallel to CE, meeting EB in F (I. 31); through F draw FG parallel to BA, and join AF. Demonstration. Then, because AC is equal to CE, therefore

1. The angle AEC is equal to the angle EAC (I. 5);

and because ACE is a right angle, therefore

2. The two angles AEC, EAC are together equal to a right

angle (I. 32);

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and since they are equal to one another, therefore

3. Each of the angles AEC, EAC is half a right angle. For the same reason,

4. Each of the angles CEB, EBC is half a right angle; and therefore

5. 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 (I. 29), therefore the remaining angle EFG is half a right angle; wherefore

6. The angle GEF is equal to the angle EFG, and the side

GF is equal to the side EG (I. 6). Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal to the interior and opposite angle. ECB (I. 29), therefore the remaining angle BFD is half a right angle; wherefore

7. The angle at B is equal to the angle BFD, and the side DF

is equal to the side DB (I. 6). And because AC is equal to CE, the square on AC is equal to the square on CE; therefore

8. The squares on AC, CE are double of the square on AC; but the square on AE is equal to the squares on AC, CE (I. 47), because ACE is a right angle; therefore

9. The square on AE is double of the square on AC. Again, because EG is equal to GF, the square on EG is equal to the square on GF; therefore

10. The squares on EG, GF are double of the square on GF; but the square on EF is equal to the squares on EG, GF (I. 47); therefore

11. The square on EF is double of the square on GF; and GF is equal to CD (I. 34); therefore

12. The square on EF is double of the square on CD; but the square on AE is double of the square on AC; therefore

13. The squares on AE, EF are double of the squares on AC,

CD; but the square on AF is equal to the squares on AE, EF, because AEF is a right angle (I. 47); therefore

14. The square on AF is double of the squares on AC, CD; but the squares on AD, DF are equal to the square on AF, because the angle ADF is a right angle (I. 47); therefore

15. The squares on AD, DF are double of the squares on AC,

CD;

and DF is equal to DB; therefore

16. The squares on AD, DB are double of the squares on AC,

CD.
If therefore a straight line be divided, &c. Q.E.D.

PROPOSITION 10.-Theorem.

If a straight line be bisected, and produced to any point, the square on the vhole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected, and of the square on the line 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 squares on AD, DB, shall be double of the squares on AC, CD.

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B

Construction. From the point C draw CE at right angles to AB (I. 11); make CE equal to AC or CB (1. 3), and join AE, EB; through E draw EF parallel to AB (I. 31); and through D draw DF parallel to CE, meeting EF in F. Then, because the straight

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