Demonstration. Then because the complement CH is equal to the complement HF (I. 43), to each of these equals add DM; therefore H B 3. AL is equal to DF; to each of these equals add CH, and therefore M 1. The whole CM is equal to the whole DF; but because the line AC is equal to CB, therefore 2. AL is equal to CM (I. 36); therefore also DB: but AH is the rectangle contained by AD, DB; for DH is equal to DB, and DF together with CH is the gnomon CMG; therefore 5. The gnomon CMG is equal to the rectangle AD, to each of these equals add LG, which is equal to the square on CD (II. 4. Cor.), therefore 6. The gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square on CD ; but the gnomon CMG and LG make up the whole figure CEFB, which is the square on CB; therefore 7. The rectangle AD, DB, together with the square on CD, is equal to the square on CB. Wherefore, if a straight line, &c. Q.E.D. Cor. From this proposition it is manifest that the difference of the squares on two unequal lines AC. CD, is equal to the rectangle contained by their sum (AD) and their difference (DB). PROPOSITION 6.-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 on half the line bisected, is equal to the square on 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. Then the rectangle AD, DB, together with the square on CB, shall be equal to the square on CD. fore B M E G F Construction. Upon CD describe the square CEFD (I. 46), and join DE; through B draw BHG parallel to CE or DF (I. 31), meeting DE in H, and EF in G; through H draw KLM parallel to AD or EF, meeting DF in M, and CE in L; and through A draw AK parallel to CL or DM, meeting MLK in K. Demonstration. Then because the line AC is equal to CB, there 1. The rectangle AL is equal to the rectangle (I. 36), but CH is equal to HF (I. 43), therefore 2. AL is equal to HF; to each of these equals 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 to DB (II. 4. Cor.), therefore 4. The gnomon CMG is equal to the rectangle AD, DB; to each of these equals add LG, which is equal to the square on CB, therefore is 5. The rectangle AD, DB, together with the square on CB, 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 on CD, therefore 6. The rectangle AD, DB, together with the square on CB, is equal to the square on CD. Wherefore, if a straight line, &c. Q.E.D. PROPOSITION 7.-Theorem. If a straight line be divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part. Let the straight line AB be divided into any two parts in the point C. Then the squares on AB, BC shall be equal to twice the rectangle AB, BC, together with the square on AC. A H and therefore D but C E F Construction. Upon AB describe the square ADEB (I. 46), and join BD; through draw CF parallel to AD or BE (I. 31), meeting BD in G, and DE in F; through G draw HGK parallel to AB or DE, meeting AD in H, and BE in K. B Demonstration. Then because AG is equal to GE (I. 43), add to each of them CK; therefore 1. The whole AK is equal to the whole CE, K 2. AK, CE are double of AK; but AK, CE are the gnomon AKF and the square CK; therefore 3. The gnomon AKF and the square CK are double of AK; for BK is equal to BC (II. 4. Cor.), therefore 4. Twice the rectangle AB, BC, is double of AK; 5. The gnomon AKF and the square CK are equal to twice the rectangle AB, BC: to each of these equals add HF, which is equal to the square on AC; therefore 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. M 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. A X E for the same reason, B D N H F 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; 2. PR is equal to RO; and because CB is equal to BD, and GK to KN, therefore A C B and therefore Σ X therefore D N E H L 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; 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; 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; |