PROP. I. THEOR. If there be iwo straight lines, one of which is divided into any number of parts ; the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines ; and let BC be divided into any parts in the points D, E; the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC. From the point B draw (11. 1.) BF at right angles to BC, and make BG equal (3. 1.) to A ; and B D E С through G draw (31. 1.) GH parallel to BC, and through D, E, C, draw (31. 1.) DK, EL, CH parallel to BG ; then BH, BK, DL, and EH are rectangles, and BH = BK G of DL + EH. K L HD But BH =BG.BC = A.BC, because BG = A: Also BK =- BG. IT" A BD = A.BD, because BG = and DL = DK.DE = A.DE, because (34. 1.) DK = BG = A. In like manner, EH = A.EC. Therefore A.BČ A.BD + A.DE + A.EC ; that is, the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC. Therefore, if there be two straight lines, &c. Q. E. D. A ; PROP. II. THEOR. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts are together equal to the square of the whole line. Let the straight line AB be divided into any two parts in the point C; the rectangle AB.BC, together with the rect A C B angle AB.AC, is equal to the square of AB; or AB.AC to AB.BC = AB. AD.AC AB.AC, because AD D F E AB; 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 foresaid part. Let the straight line AB be divided into two parts in the point C; the rectangle AB.BC is equal to the rectangle AC.BC, together with BC2. Upon BC describe (46. 1.) the A square CDEB, and produce ÉD to F, and through A draw (31. 1:) AF parallel to CD or BE ; then AÉ AD + CE. But AE = AB.BE = because BE = BC. So also ADAC.CD AC.CB; and CE BC2 ; therefore AB.BC = AC.CB + BC2. Therefore, if a straight F line, &c. Q. E. D. AB.BC, PROP. 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, that is, AB2 = ACS + CB2 + 2AC.CB. Upon AB describe (46. 1.) the square ADEB, and join BD, and through C draw (31. 1.) CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (29. 1.) to the interior A С B and opposite angle ADB ; but ADB is equal (5. 1.) to the angle ABD, because G BA is equal to AD, being sides of a square; H K wherefore the angle CGB is equal to the angle GBC ; and therefore the side BC is equal (6. 1.) to the side CG : but CB is equal (34. 1.) also to GK, and CG to BK ; wherefore the figure CGKB is equilateral. D E It is likewise rectangular ; for the angle CBK being a right angle, the other angles of the parallelogram CGKB are also right angles (Cor. 46. 1.). Wherefore CGKB is a square, it is upon the side CB. For the same reason HF also is a square, and / 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 (43. 1.) to the complement GE ; and because AG=AC.CG=AC.CB, therefore also GE=AC.CB, and AG+GE=2AC.CB. Now, HFACP and CK=CB2 ; therefore, HF+CK+AG+GE=AC: +CB2 + 2AC.CB. But HF+CK+AGGE=the figure AE, or AB; therefore AB2 =AC: +CB2 +2AC.CB. Wherefore, if a straight line be divided &c. Q. E. D. Cor. From the demonstration, it is manifest that the parallelograms about the diameter of a square are likewise squares. PROP. V. THEOR. If a straight line be divided into two equal parts, and also into two un equal 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 in the point D; the rectangle AD.DB, together with the square of CD, is equal to the square of CB, or AD.DB+CD=CB2. Upon CB describe (46. 1.) the square CEFB, join BE, and through D draw (31. 1.) DHG parallel to CE or BF ; and through H draw KLM parallel to CB or EF ; and also through A draw AK parallel A C D to CL or BM: And because CH =HF, if DM be added to both, CM=DF. But AL=(36.1.) CM, KY H AD.DH AD.DB, because DH - DB (Cor. 4. 2.); E 6 Ꮐ therefore gnomon CMG AD.DB. To each add LG then gnoman CMG + LG AD.DB to CD2 ,But CMG + LG therefore AD.DB+CD=BC. Wherefore, if a straight line, &c. R. 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, or that AC2-CD2=(ACT CD) (AC-CD)." PROP. 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 M CD', BC2! D; the rectangle AD.DB, together with the square of CB, is equal to the square of ČD. Upon CD describe (46. 1.) the square CEFD, join DE, and through B draw (31. 1.) BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK parallel to CL or DM. And because AC is equal to CB, the rectangle AL is equal (36. 1.) to CH > but CH is equal (43. 1.) to HF; therefore also AL is equal A С B D to HF: To each of these add CM; therefore the whole AM is equal to the gnomon CMG. Now K L H M gnomon CMG=AD.DB, and CMG+LG =AD.DB + CB2. But CMG+ LG= CF = CD2, therefore AD. E G F PROP. VII. THEOR. line, and of one of the parts, are equal to twice the rectangle con- 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, or AB? +BC2=2AB.BC+AC2. Upon AB describe (46. 1.) the square ADEB, and construct the figure as in the preceding propositions : Because AG=GE, AG+CK =GE+CK, that is, AK=CE, and therefore AK+CE=2AK. But AK + CE= A C B gnomon AKF+CK ; and therefore, AKF +CK = 2AK = 2AB.BK=2AB.BC, be G. cause BK = (Cor. 4. 2.) BC. Since then, H K AKF+CK=2AB.BC, AKF + CK + HF = 2AB.BC + HF; and because AKF + HF=AE=AB2, ABS +CK = 2AB.BC+ HF, that is, (since CK = CBP, and HF - AC2,) AB? + CB2= 2AB.BC + AC2. D E Wherefore, if a straight line, &c. Q. E. D. Otherwise, “Because AB = AC + BC2 + 2AC.BC (4. 2.), adding BC to "both, AB2 +-BC2=AC: +2BC: +2AC.BC. с " But BC? + AC.BC = AB.BC (3. 2.); and A B therefore, 2BC: +2AC.BC=2ÀB.BC ; and "therefore ABS + BCP=AC: +2AB.BC." H “ Cor. Hence, the sum of the squares of any two lines is equal " to twice the rectangle contained by the lines together with the square of the difference of the lines.? 66 PROP. 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 up of the whole and the first-mentioned 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 GK is equal (34. 1.) to CB, and CB to BD, and BD to KŇ, 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 rectangles CK and BN are equal, as also the rectangles GR and RN: But CK is equal (43. 1.) to RN, because they are the complements of the 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 CK+BN+GR+RN=4CK. Again, because CB is equal to BD, and BD equal (Cor. 4. 2.) to BK, that A is, to CG ; and CB equal to GK, that с в D (Cor. 4. 2.) is, to GP; therefore CG is equal to GP; and because CG is G K. equal to GP, and PR to RO, the rect N angle AG is equal to MP, and PL to P Р R 0 RF : But MP is equal (43. 1.) 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, E are equal to one another, and so AG H. L 1 +MP+PL+RF=4AG. And it was demonstrated, that CK+BN+GR+RN=4CK; wherefore, adding equals to equals, the whole gnomon AOH=4AK. Now AK=AB.BK =AB.BC, and 4 AK=4AB.BC ; therefore, gnomon AOH=4AB.BC; and adding XH, or (Cor, 4. 2.) AC?, to both, gnomon AOH+XH= 4AB.BC FAC?. But AOH + XH=AF = AD2 ; therefore AD>= 4A.BBC+AC2. Now AD is the line that is made up of AB and BC, added together into one liue : Wherefore, if a straight line, &c. Q. E. D. " Cor. 1. Hence, because AD is the sum, and AC the difference of " the lines AB and BC, four times the rectangle contained by any two " lines together with the square of their difference, is equal to the square of the sum of the lines," |