10. And that BD is equal to BK, that is CG, and I. 34.) 11. And that CB is equal to GK, that is GP, II. 4, cor.) 12. Therefore CG is equal to GP._(ax. 1.) (II. 4, cor., (I. 34, and 13. And because CG is equal to GP, and PR to RO, 14. The rectangle AG is equal to MP, and PL to RF. (I. 36.) 15. But MP is equal to PL, because they are complements of the parallelogram ML (I. 43), and AG is equal to RF._(ax. 1.) 16. Therefore the four rectangles AG, MP, PL, RF, are equal to one another, and so the four are quadruple of one of them, AG. 17. And it was demonstrated that the four CK, BN, GR, and RN, are quadruple of CK. (dem. 8.) 18. Therefore the eight rectangles which make up the gnomon AOH are quadruple of AK. 19. And because AK is the rectangle contained by AB and BC, for BK is equal to BC, 20. Therefore four times the rectangle AB BC is quadruple of AK. 21. But the gnomon AOH was demonstrated to be quadruple of AK. (dem. 18.) 22. Therefore four times the rectangle AB BC is equal to the gnomon AOH. (ax. 1.) 23. To each of these add XH, which is equal to the square on AC. (II. 4, cor., and I. 34.) 24. Therefore four times the rectangle AB BC, together with the square on AC, is equal to the gnomon AOH, and the square XH. 25. But the gnomon AOH and the square XH make up the figure AEFD, which is the square on AD. 26. Therefore four times the rectangle AB BC, together with the square on AC, is equal to the square on AD, that is, on the line made up of AB and BC together. Conclusion. Therefore, 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. (References-Prop. I. 3, 5, 6, 11, 29, 31, 32, 34, 47.) Hypothesis. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D. Sequence. The squares on AD and DB, shall be together double of the squares on AC and CD. Construction.-1. From the point C draw CE at right angles to AB (I. 11), and make it equal to AC or CB (I. 3), and join EA, EB. 2. Through D draw DF parallel to CE. (I. 31.) 3. Through F draw FG parallel to BA (I. 31), and join AF. Demonstration.-1. Because AC E G F C D B is equal to CE (const.), the angle EAC is equal to the angle AEC. (I. 5.) 2. And because the angle ACE is a right angle (const.), the angles AEC and EAC together make one right angle (Í. 32), and they are equal to one another. 3. Therefore each of the angles AEC and EAC is half a right angle. 4. For the same reason, each of the angles CEB and EBC is half a right angle. 5. Therefore the whole angle AEB is a right angle. 6. And because the angle GEF is half a right angle, and the angle EGF a right angle, for it is equal to the interior and opposite angle ECB, (I. 29.) 7. Therefore the remaining angle EFG is half a right angle. 8. Therefore the angle GEF is equal to the angle EFG, and the side EG is equal to the side GF. (I. 6.) 9. Again, because the angle at B is half a right angle, and the angle FDB a right angle, for it is equal to the interior and opposite angle ECB, (I. 29.) 10. Therefore the remaining angle BFD is half a right angle. 11. Therefore the angle at B is equal to the angle BFD, and the side DF is equal to the side DB. (I. 6.) 12. And because AC is equal to CE (const.), the square on AC is equal to the square on CE. 13. Therefore the squares on AC and CE are double of the square on AC. 14. But the square on AE is equal to the squares on AC and CE, because the angle ACE is a right angle; (I. 47.) 15. Therefore the square on AE is double of the square on AC. 16. Again, because EG is equal to GF (const.), the square on EG is equal to the square on GF; 17. Therefore the squares on EG and GF are double of the square on GF. 18. But the square on EF is equal to the squares on EG and GF, because the angle EGF is a right angle; (I. 47.) 19. Therefore the square on EF is double of the square on GF. 20. And GF is equal to CD, (I. 34.) 21. Therefore the square on EF is double of the square on CD. 22. But it has been demonstrated that the square on AE is also double of the square on AC;_(dem. 15.) 23. Therefore the squares on AE and EF are double of the squares on AC and CD. 24. But the square on AF is equal to the squares on AE and EF, because the angle AEF is a right angle; (I. 47.) 25. Therefore the square on AF is double of the squares on AC and CD. 26. But the squares on AD and DF are equal to the square on AF, because the angle ADF is a right angle; (I. 47.) 27. Therefore the squares on AD and DF are double of the squares on AC and CD. 28. And DF is equal to DB (dem. 11); therefore the squares on AD and DB are double of the squares on AC and CL. Conclusion. Therefore, if a straight line, &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. (References-Prop. I. 5, 6, 11, 15, 29, 31, 32, 46, 47.) Hypothesis. Let the straight line AB be bisected in C, and produced to D. Sequence. The squares on AD and DB shall be together double of the squares on AC and CD. Construction.-1. From the point C draw CE at right angles to AB, and make it equal to AC or CB (I. 11, I. 3), and join AE, EB. 2. Through E draw EF parallel to AB, and through D draw DF parallel to CE (I. 31.) Then because the straight line EF meets the parallels EC, FD, the angles CEF, EFD, are equal to two right angles (I. 29); therefore the angles BEF, EFD, are less than two right angles; therefore EB, FD, will meet, if produced towards B and D. (ax. 12.) 3. Let them meet in G, and join AG. Demonstration.-1. Because AC is equal to CE (const.), the angle CEA is equal to the angle EAC. (I. 5.) 2. And the angle ACE is a right angle, therefore each of the angles CEA and EAC is half a right angle. (I. 32.) 3. For the same reason each of the angles CEB and EBC is half a right angle. 4. Therefore the whole angle AEB is a right angle. 5. And because the angle EBC is half a right angle, the angle DBG, which is vertically opposite, is also half a right angle; (I. 15.) 6. But the angle BDG is a right angle, because it is equal to the alternate angle DCE; (I. 29.) 7. Therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; 8. Therefore also the side BD is equal to the side DG. (I. 6.) 9. Again, because the angle EGF is half a right angle, and the angle at F a right angle, for it is equal to the opposite angle ECD; (I. 34.) 10. Therefore the remaining angle FEG is half a right angle (I. 32), and therefore equal to the angle EGF; 11. Therefore also the side GF is equal to the side FE. (I.6.) 12. And because EC is equal to CA, the square on EC is equal to the square on CA; 13. Therefore the squares on EC and CA are double of the square on CA. 14. But the square on AE is equal to the squares on EC and CA; (I. 47.) 15. Therefore the square on AE is double of the square on AC. 16. Again, because GF is equal to FE, the square on GF is equal to the square on FE; 17. Therefore the squares on GF and FE are double of the square on FE. 18. But the square on EG is equal to the squares on GF and FE; (I. 47.) 19. Therefore the square on EG is double of the square on FE. 20. And FE is equal to CD, (I. 34.) 21. Therefore the square on EG is double of the square on CD. 22. But it has been demonstrated that the square on AE is double of the square on AC; (dem. 15.) 23. Therefore the squares on AE and EG are double of the squares on AC and CD. 24. But the square on AG is equal to the squares on AE and EG; (I. 47.) 25. Therefore the square on AG is double of the squares on AC and CD. 26. But the squares on AD and DG are equal to the square on AG; (I. 47.) 27. Therefore the squares on AD and DG are double of the squares on AC and CD. 28. And DG is equal to DB (dem. 8); therefore the squares on AD and DB are double of the squares on AC and CD. Conclusion. Therefore, 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. (References-Prop. I. 3, 10, 46, 47; II. 6.) Given.-Let AB be the given straight line. Sought. It is required to divide AB 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. Construction.-1. Upon AB de scribe the square ABDC. (I. 46.) 3. Produce CA to F, and make EF equal to EB. (I. 3.) 4. Upon AF describe the square AFGH. (I. 46.) 5. Produce GH to K. Then AB shall be divided in II, so that the rectangle AB BH is equal to the square on AH. Proof.-1. Because the straight line AC is bisected in E, and produced to F. F G H A B E 2. The rectangle CF. FA, together with the square on AE, is equal to the square on EF. (II. 6.) 3. But EF is equal to EB, (const.) 4. Therefore the rectangle CF. FA, together with the square on AE, is equal to the square on EB. |
