The Elements of Euclid, containing the first six books, with a selection of geometrical problems. To which is added the parts of the eleventh and twelfth books which are usually read at the universities. By J. Martin

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

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

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.

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);

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 whole 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.

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 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

line EF meets the parallels CE, FD, therefore the angles CEF, EFD are equal to two right angles (I. 29), and therefore

1. The angles BEF, EFD are less than two right angles.

But straight lines which, with another straight line, make the interior angles upon the same side of a line less than two right angles, will meet if produced far enough (I. Ax. 12); therefore

2. EB, FD will meet, if produced towards B, D;

let them be produced and meet in G, and join AG.

Demonstration. Then, because AC is equal to CE, therefore the angle CEA is equal to the angle EAC (I. 5), and the angle ACE is a right angle, therefore

3. Each of the angles CEA, EAC is half a right angle (I. 32). For the same reason,

therefore

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

5. The whole AEB is a right angle.

And because EBC is half a right angle, therefore also 6. DBG is half a right angle (I. 15),

for they are vertically opposite; but

7. BDG is a right angle,

because it is equal to the alternate angle DCE (I. 29); therefore 8. The remaining angle DGB is half a right angle;

and it is therefore equal to the angle DBG; wherefore also 9. The side BD is equal to the side DG (I. 6).

Again, because EGF is half a right angle, and the angle at Fis a right angle, being equal to the opposite angle ECD (I. 34), therefore 10. The angle FEG is half a right angle, and equal to the angle EGF;

wherefore also

11. The side GF is equal to the side FE (I. 6).

And because EC is equal to CA, the square on EC is equal to the square on CA; therefore

12. The squares on EC, CA are double of the square on CA; but the square on EA is equal to the squares on EC, CA (I. 47), therefore

13. The square on EA is double of the square on AC. Again, because GF is equal to FE, the square on GF is equal to the square on FE; therefore

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

15. The square on EG is double of the square on FE;

and FE is equal to CD (I. 34); wherefore

16. The square on EG is double of the square on CD;

but it was demonstrated, that the square on EA is double of the square on AC, therefore

17. The squares on EA, EG are double of the squares on AC,

CD;

but the square on AG is equal to the squares on EA, EG (I. 47); therefore

18. The square on AG is double af the squares on AC, CD; but the squares on AD, DG are equal to the square on AG, therefore 19. The squares on AD, DG are double of the squares on AC,

CD;

but DG is equal to DB; therefore

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

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

PROPOSITION 11.-Problem.

To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square on the other part.

Let AB be the given straight line. It is required to divide AB into two parts, so that the rectangle contained by the whole line, and one of the parts, shall be equal to the square on the other part.

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The Elements of Euclid, containing the first six books, with a selection of ...