D E C make BG equal (I. 3) to, A; and through G draw (I. 31) GH G F K L H A COR. The rectangle contained by one straight line and a part of another is equal to the difference of the rectangles contained by the undivided line, and the whole and remaining part of the divided line. 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. C B Let the straight line AB be divided into any two parts in the point C; the rectangle AB.BC, together with the A rectangle AB.AC, is equal to the square of AB; or AB.AC+AB.BC=AB2. On AB describe (I. 46) the square ADEB, and through C draw CF (I. 31) parallel to AD or BE; then AF+CE=AE; but AF=AD.AC=AB.AC, because AD-AB; CE=BE.BC=AB.BC; and AE= AB2; therefore AB.AC+AB.BC= AB. D Therefore, if a straight line, &c. Q. E. D. PROP. III. THEOR. F E 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 any two parts in the point C; the rectangle AB.BC is equal to the rectangle AC.BC, together with BC2. Upon BC describe (I. 46) the square CDEB, and produce ED to F, and through A draw (I. 31) AF parallel to CD or BE; then AE-AD+CE; but AE=AB.BE= AB.BC, because BE-BC; so also AD= AC.CD=AC.CB; and CE=BC2; therefore AB BC=AC.CB+BC2. Therefore, if a straight line, &c. Q E. D. A C B D E 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=AC2+CB2+ 2AC.CB. A H D 1=2(5) B 4 G 3 K 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 HK parallel to AB or DE; and because CF is paralle to AD, and BD falls upon them, the exterior angle BGC is equal (I. 29) to the interior and opposite angle ADB; but ADB is equal (I. 5) to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle GBC; and therefore the side BC is equal (I. 6) to the side CG; but CB is equal (I. 34) also to GK, and CG to BK; wherefore the figure CGKB is equilateral. It is likewise rectangular; for the angle CBK being a right angle, the other angles of the parallelogram CGKB are also right angles (I. 46, Cor.); wherefore CGKB 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 (I. 43) to the complement GE; and because AG-AC.CG-AC.CB, therefore also, GE=AC.CB, and AG+GE=2AC.CB; now, HFAC2, and CK=CB2; therefore HF+CK+AG+GE=AC2+CB2+2AC.CB. But HF+CK+AG÷GE=the figure AE, or AB2; therefore AB-AC2+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 diagonal of a square are likewise squares. PROP. V. THEOR. If a straight line be divided into two equal parts, and also intor 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 in the point D; the rectangle AD.DB, together with the square of CD, is equal to the square of CB, or AD,DB+CD2=CB2. Upon CB describe (I. 46) the square CEFB; join BE; and through D draw (I. 31) 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 CH-HF (1. 43), if DM C A C D B be added to both, CM=DF; but AL=(I. 36) CM; therefore AL= DF, and adding CH to both, AH=gnomon CMG; but AH=AD. DH=AD.DB, because DH=DB (II. 4, Cor.); therefore gnomon CMG=AD.DB; to each add LG =CD2, then gnomon CMG+LG= & AD.DB+CD2; but CMG+LG= BC2; therefore AD.DB+CD2= BC. Wherefore, if a straight line, &c. Q. E. D. M H E G F 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= (AC+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 D; the rectangle AD.DB, together with the square of CB, is equal to the square of CD. C B D M L H 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 AK parallel to CL or DM; and be- A cause AC is equal to CB, the rectangle AL is equal (I. 36) to CH; but CH is equal (I. 43) to HF; k therefore also, AL is equal to HF; to each of these add CM; therefore the whole AM is equal to the gnomon CMG; now, AM=AD.DM=AD.DB, because DM (II. 4, Cor.)=DB; therefore gnomon CMG-AD.DB, and CMG+LG=AD.DB+CB2; but CMG+LG=CF=CD2; therefore AD.DB+CB2=CD2. Therefore, if a straight line, &c. Q. E. D. PROP. VII. THEOR. E G F 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, or AB2+BC2=2AB.BC+ AC2. B K Upon AB describe (I. 46) the square ADEB, and construct the figure as in the preceding propositions; because AG=GE (I. 43) AG+CK=GE+CK, that is, ÁK=CE; and therefore AK+CE=2AK; but AK+CE: A gnomon AKF+CK; and therefore AKF+ CK=2AK=2AB.BK=2AB.BC, because BK H =(II. 4, Cor.) BC; since then, AKF÷CK= 2 AB.BC, AKF÷CK+HF=2AB.BC+HF; and because AKF+HF=AE=AB2, AB2+ CK=2AB.BC+HF; that is (since CK=CB2. and HF-AC), AB2+CB2=2AB.BC+AC2. Wherefore, if a straight line, &c. Q. E. B. Otherwise: D E F Because AB2-AC2+BC2+2AC.BC (II. 4) adding BC2 to both AB2+BC2= AC2+2BC2 +2AC.BC; but BC2+AC.BC=AB.BC (II. 3); and therefore, 2BC2+2AC.BC=2AB.BC; and therefore AB2+BC2=AC2+2AB BC. 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. 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 firstmentioned 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 (I. 34) to CB, and CB to BD, and BD to KN, GK is equal to KN. For the same reason, PR is equal to RO; and because CB is equal to BD, and KG to KN, the rectangles CK and BN are equal, as also the rectangles GR and RN; but CK is equal (I. 43) 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 (II. 4, Cor.) to BK, M that is, to CG; and CB equal to GK, that (II. 4. Cor.) 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; but MP is equal (I. 43) to PL, because they are the complements of the E parallelogram ML; wherefore AG is equal also to RF; there X A C B D G K N P R H L F fore the four rectangles AG, MP, PL, RF are equal to one another, and so AG+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 4AK=4AB.BC; therefore gnomon AOH=4AB.BC; and adding XH or (II. 4, Cor.) AC2 to both, gnomon AOH+XH =4AB BC+AC2; but AOH+XH=AF=AD2; therefore AD2= 4AB.BC+AC2. Now AD is the line that is made up of AB and BC, added together into one line. 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. COR. 2. From the demonstration it is manifest, that since the square of CD is quadruple of the square of CB; the square of any line is quadruple of the square of half that line. Otherwise: Because AD is divided anyhow in C (II. 4) AD2=AC2+CD2+ 2CD.AC; but CD=2CB; and therefore CD2=CB2+BD2+2CB. BD (II. 4)=4CB2; and also 2CD.AC-4CB.AC; therefore AD2= AC2+4BC2+4BC.AC; now BC2+BC.AC = AB.BC (II. 3); and therefore AD AC2+4AB.BC. Q. E.D. PROP. IX. THEOR. 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 C into two equal, and at D into two unequal parts; the squares of AD, DB are together double the squares of AC, CD. E From the point C draw (I. 11) CE at right angles to AB, and make it equal 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, the angle EAC is equal (I. 5) to the angle AEC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angle (I. 32); and they are equal to one another; each of them therefore is half a right angle. For the same reason, each of the angles CEB, EBC is half a right angle; and therefore 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 opposite angle ECB, the remaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the side EG equal A D F |