PROPOSITION III. THEOR. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part. Let the straight line AB be divided into any two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. Upon BC describe* the square CDEB, and produce ED 46. 1. to F, and through A drawf AF parallel to CD or BE; then the rectangle AE is equal to the rectangles AD, CE; And AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is F equal to BC; с D E B And AD is contained by AC, CB, for CD is equal to CB; And DB is the square of BC; Therefore the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. If therefore a straight line, &c. Q. E. D. PROPOSITION IV. THEOR. If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB. + 31. 1. + 31. 1. с B Upon AB describe* the square ADEB, and join BD, * 46. 1. and through C draw† CGF parallel to AD or BE, and through G draw HK parallel to AB or DE: And because CF is parallel to AD, H and BD falls upon them, the exterior angle BGC is equal to the interior and opposite angle ADB; but ADB is equal to the angle ABD, because BA D K † 29. 1. F E || 5. 1. is equals to AD, being sides of a square; wherefore the § 30 Def. 1 Ax. * 6. 1. 34. 1. 34. 1. § 43. 1. 46. 1. + 31. 1. * angle CGB is equal ¶ to the angle GBC; and therefore the side BC is equal to the side CG: but CB is equal† also to GK, and CG to BK; wherefore the figure CGKB is equilateral : It is likewise rectangular; for CG is parallel to BK, and CB meets them; the angles KBC, GCB are therefore equal to two right angles; and KBC is a right angle; wherefore GCB is a right angle: and therefore also the angles‡ CGK, GKB opposite to these, are right angles, and CGKB is rectangular; but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB: For the same reason HF also is a square, and it is upon the side HG, which is equal to AC: therefore HF, CK are the squares of AC, CB: And because the complement AG is equal§ to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB: And HF, CK are the squares of AC, CB; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB: But HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore, if a straight line, &c. Q. E. D. COR. From the demonstration, it is manifest, that the parallelograms about the diameter of a square are likewise squares. PROPOSITION V. THEOR. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D; the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Upon CB describe* the square CEFB, join BE, and through D draw† DHG parallel to CE or BF; and through H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM: And because the complement CH is equal to the com- 43. 1. plement HF, to each of these add A DM; therefore the whole CM is equal to the whole DF; but CM is equal to AL, because AC is equal to CB; therefore also AL is equal to DF. To each of these add CH, and K the whole AH is equal to DF and CH: с D B L H 2 Ax. M $ 36. 1. E G F But AH is the rectangle contained by AD, DB, for DH is equal¶ to DB; and DF together with CH is the gnomon ¶ Cor. 4. 2. CMG; Therefore the gnomon CMG is equal to the rectangle AD, DB: To each of these add LG, which is equal to the square of CD; Therefore 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 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. 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. PROPOSITION VI. THEOR.-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. Upon CD describe the square CEFD, join DE, and through B draw† BHG parallel to A CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK parallel K to CL or DM; And because AC is equal to CB, the rectangle AL is equal to CH; • Cor. 4. 2. 46. 1. с BD +31. 1. but CH is equal to HF; therefore also AL is equal to HF : || 43. 1. To each of these add CM; therefore the whole AM is equal to the gnomon CMG: And AM is the rectangle contained by AD, DB, for Cor. 4.2. DM is equal§ to DB: therefore the gnomon CMG is equal to the rectangle AD, DB: 46. 1. + 43. 1. Add to each of these LG, which is equal to the square of CB; therefore 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 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. PROPOSITION VII. THEOR.-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. 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. Upon AB describe* the square ADEB, and construct the figure as in the preceding propositions; And because AG is equal to GE, add to each of them CK; the whole AK is therefore equal to the whole CE; therefore AK, CE are double of AK: But AK, CE are the gnomon AKF, together with the square CK; therefore the gnomon AKF, together with the square CK, is double of AK: But twice the rectangle AB, BC is H G K F Cor. 4.2. double of AK, for BK is equal to BC; therefore 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 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 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. PROPOSITION VIII. THEOR.-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 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. Produce AB to D, so that BD be equal to CB, and upon AD describe the square AEFD; and construct two figures such as in the preceding. Because CB is equal to BD, and that CB is equal* to * 34. 1. GK, and BD to KN; therefore GK is equal to KN: For the same reason, PR is equal to RO; And because CB is equal to BD, and GK to KN, the rectangle CK is equal† to BN, and GR to RN; but CK is + 36. 1. equal to RN, because they are the complements of the ‡ 43. 1. parallelogram CO; therefore also BN is equal to GR; and the four rectangles BN, CK, GR, RN are therefore 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|| Cor. 4.2. to BK, that is, to CG, and CB equal to GK, that is, to GP; therefore CG is equal to GP: And because CG is equal to GP, and PR to RO, the rectangle AG is equal to MP, and PL to RF: A CBD M G K N P R x E HLF $ 43. 1. But MP is equal§ to PL, because they are the complements of the parallelogram ML: wherefore AG is equal also to RF: therefore 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 the eight rectangles which contain the gnomon AOH, are quadruple of AK; And because AK is the rectangle contained by AB, BC, |