But AH is the rectangle contained by AD, BD, for DH is equal (II. 4. Cor.) to DB; and DF, together with CH, is the gnomon CMG; therefore 4. The gnomon CMG is equal to the rectangle AD, DB: to each of these add LG, which is equal (II. 4. Cor.) to the square of CD; therefore 5. The gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square of CD: but the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB; therefore 6. The rectangle AD, DB, together with the square of CD, is equal to the square of CB. Wherefore, if a straight line, &c. Q.E.D. COR. From this proposition it is manifest, that the difference of the squares of two unequal lines, AC, CD, is equal to the rectangle contained by their sum and difference. PROP. VI. -THEOREM. If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of` the straight line which is made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to the point D; the rectangle AD, DB, together with the square of CB, is equal to the square of CD. B D H K с L M E G F Upon CD describe (I. 46.) the square CEFD, join DE, and through B draw (I. 31.) BHG parallel to CE or DF; and through H draw KLM parallel to AD or EF; and also through A draw 4K parallel to CL or DM. And because AC is equal (Hyp.) to CB, (I. 36.) 1. The rectangle AL is equal to CH; but (I. 43.) CH is equal to HF; therefore also 2. AL is equal to HF: to each of these add CM; therefore 3. The whole AM is equal to the gnomon CMG: but AM is the rectangle contained by AD, DB, for DM is equal (II. 4. Cor.) to DB; therefore 4. The gnomon CMG is equal to the rectangle AD, DB: add to each of these LG, which is equal to the square of CB; therefore 5. The rectangle AD, DB, together with the square of CB, is equal to the gnomon CMG and the figure LG. But the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore 6. The rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c. Q.E.D. PROP. VII. THEOREM. If a straight line be divided into any two parts, the squares of the whole line and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. but Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC, are equal to twice the rectangle AB, BC, together with the square of AC. C B G H Z D F E Upon AB describe (I. 46.) the square ADEB, and join BD, and through C draw (I. 31.) CGF parallel to AD or BE, and through G draw HGK parallel to AB or DE. And because AG is equal (I. 43.) to GE, add to each of them CK; therefore 1. The whole AK is equal to the whole CE; and therefore 2. AK, CE, are double of AK; but AK, CE, are the gnomon AKF, together with the square CK; therefore K 3. The gnomon AKF, together with the square CK, is double of AK; 4. Twice the rectangle AB, BC, is double of AK, for BK is equal (II. 4. Cor.) to BC; therefore 5. The gnomon AKF, together with the square CK, is equal to twice the rectangle AB, BC; to each of these equals add HF, which is equal to the square of AC; therefore 6. The gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC, and the square of AC; but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC; therefore 7. The squares of AB and BC are equal to twice the rectangle AB, BC, together with the square of AC. Wherefore, if a straight line, &c. Q.E.D. PROP. VIII.—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 of the other part, is equal to the square of 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; four times the rectangle AB, BC, together with the square of AC, is equal to the square of the straight line made up of AB and BC together. A CBD M X for the same reason, 2. GK P R E HL F Produce AB to D, so that BD be equal (I. 3.) to CB; and upon AD describe (I. 46.) the square AEFD; and construct two figures such as in the preceding. Because CB is equal to BD, and that CB is equal (I. 34.) to GK, and BD to KN; therefore 1. GK is equal to KN; N 0 PR is equal to RO; and because CB is equal to BD, and GK to KN, (I. 36.) 3. The rectangle CK is equal to BN, and GR to RN; but CK is equal (I. 43.) to RN, 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 that BD is equal (II. 4. Cor.) to BK, that is (I. 34.) to CG; and CB 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, 7. The rectangle AG is equal to MP, and PL to RF; but MP is equal (I. 43.) to PL, 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 AK: 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 40H 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 add XH, which is equal (II. 4. Cor.) to the AC; therefore square of 13. Four times the rectangle AB, BC, together with the square of AC, is equal to the gnomon AOH, and the square XH; but the gnomon 4AOH and XH make up the figure AEFD, which is the square of AD; therefore 14. Four times the rectangle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q.E.D. PROP. IX.-THEOREM. If a straight line be divided into two equal, and also into two unequal parts, the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section. Let the straight line AB be divided at the point Cinto two equal, and at D into two unequal parts; the squares of AD, DB, are together double of the squares of AC, CD. From the point C draw (I. 11.) CE at right angles to AB, and make it equal (I. 3.) to AC or CB, and join EA, EB; through D draw (I. 31.) DF parallel to CE; and through F draw FG parallel to AB; and join AF. Then, because AC is equal to CE, (I. 5.) 1. The angle EAC is equal to the angle AEC; and because the angle ACE is a right angle, (I. 32.) 2. The angles AEC, EAC, together make one right angle; and they are equal to one another; therefore 3. Each of the angles AEC, EAC, is half a right angle. For the same reason Each of the angles CEB, EBC, is half a right angle; 4. 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 (I. 29.) to the interior and opposite angle ECB, the remaining angle EFG is half a right angle; therefore 6. The angle GEF is equal to the angle EFG, and (I. 6.) the side EG is equal to the side GF. Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal (I. 29.) to the interior and opposite angle ECB, the remaining angle BFD is half a right angle; therefore 7. The angle at B is equal to the angle BFD, and (I. 6.) the side DF is equal to the side DB. And because AC is equal to CE, the square of AC is equal to the square of CE; therefore 8. The squares of AC, CE, are double of the square of AC: but the square of EA is equal (I. 47.) to the squares of AC, CE, because ACE is a right angle; therefore 9. The square of EA is double of the square of AC. Again, because EG is equal to GF, the square of EG is equal to the square of GF; therefore 10. The squares of EG, GF, are double of the square of GF; but the square of EF is equal (I. 47.) to the squares of EG, GF; therefore 11. The square of EF is double of the square of GF; and GF is equal (I. 34.) to CD; therefore 12. The square of EF is double of the square of CD; but the square of AE is likewise double of the square of AC; therefore 13. The squares of AE, EF, are double of the squares of AC, CD; |