apgle FBC, therefore the base AD is equal (4. 1.) to the base FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double (41, 1.) of the triangle ABD, because they are upon the same base, BD, and between the same parallels, BD, AL; and the square GB is double of the triangle BFC, because these also are upon the same base FB, and between the same parallels FB, GC, Now the doubles of equals are equal (6. Ax.) to one another; therefore the parallelogram BL is equal to the square GB: And in the same manner, by joining AE, BK, it is demonstrated that the parallelogram ÇL is equal to the square HC. Therefore, the whole square BDEC js equal to the two squares GB, HC; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC: wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. Therefore, in any right angled triangle, &c. , ED, PROP. XLVIII. THEOR, If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle contained by these two sides is a right angle. If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, Ibe angle BAC is a right angle, From the point A draw (11, 1.) AD at right angles to AC, and make AD equal to BĄ, and join DC. T'ben because DA is equal to AB, the quare oi DẠ is equal to the square D of AB: To each of these add the square of AC; therefore the squares of DA, AC are equal to the squares of BA, AC. But the A square of DC is equal (47. 1.) to the squares of DA, AC, because DAC is a right angle; aod the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore, the square of DC is equal to the square of BC; B and therefore also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, and the base DC likewise equal to the base BC, the aogle DAC is equal (8. 1.) to the angle BAC : But DAC is a right angle; therefore also BAC is a right angle. There fore, if the square, &c. Q. E.D. ELEMENTS OF GEOMETRY. BOOK II. DEFINITIONS. E I. contained by any two of tbe straight lines wbich are about one of the right angles. “ Thus the right angled parallelogram AC is called the rectangle "contained by AD and DC, or by AD and AB, &c. For the sake of " brevity, instead of the rectangle contuined by AD and DC, we shall “simply say the rectangle AD.DC, placing a point between the two "sides of the rectangle. Also, instead of the square of a line, for "instance of AD, we shall frequently in what follows write AD2" “ The sign + placed between the names of two magnitudes, signi“fies that those magnitudes are to be added together, and the siga placed between them, signifies that the latter is to be takea away from the former.” 66 The sign = signifies, that the things between which it is placed fare equal to one another.” II. In every parallelogram, any of А E the parallelograms about a di D ameter, together with the two complements, is called a Guo 6. Thus the parallelo“ gram HG, together with the F complements AF, FC, is the H K " gnomon of the parallelogram “ AC. This gnomoa may also, “ for the sake of brevity, be so called the goomon AGK or B .6 EHC." mon. B - BG, F A; PROP. I. THEOR. If there be two 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 DE make BG equal (3. 1.) to A; and 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 recta igles, and BH =BK G + DL + EH. K L H But BH = BG.BC A.BC, because BG=A: Also BK 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.BC : 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. 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 с B angle AB.AC, is equal to the square of AB; or AB.AC + AB.BC AB2. On AB describe (46. 1.) the square ADEB, and through C draw CF (31. 1.) parallel to AD or BE; then AF + CÈ = AÉ. But AF AD.AC AB AC, hecause AD D E PROP. III. THEOR. If a straight line be divided into any to 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; AB; the reetangle AB.BC is equal to the rectangle AC.BC, together with BC. Upon BC describe (46. 1.) the A C B square CDEB, and produce ED to F, and through A draw (31. 1.) AF parallel to ČD or BE; then AE= AD+CE. But AE = AB.BE AB.BC, because BE=BC. So also ADAC.CD = AC.CB: and CE BC%; therefore AB.BC = AC.CB +BC%. Therefore, if a straight F D line, &c. Q. E. D. 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, ABP = ACP + CBP+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 an A с в gle BGC is equal (29. 1.) to the interior and opposite angle ADB; but ADB is equal (5. 1.) to the angle ABD, because 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 It is likewise rectangular; for the angle E CBK being a right angle, the other angles of the parallelogram CGKB are also right angles (Cor. 16. 1.). 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 (43.1.) to the complement GE; and because AG SAC.CG AC.CB, therefore also GE=AC.CB, and AG+GE=2AC.CB. Now, HFAC and CK=CB'; therefore, HF+CK+AG+GE=AC FCB 2AC.CB. But HF+CK+AG+GE=the figure AE, or AB ; therefore AB =ACs +CB:+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 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+CD=CB. 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 A С D B also through A draw AK parallel to CL or BM: And because CH K! =HF, if DM be added to both, M M L |Н CM=DF, But AL=(36.1.)CM, therefere AL=DF, and adding CH to both, AH=gnomon CMG. But AH = AD.DH=AD.DB, because DH=DB (Cor. 4. 2.); E G therefore gnomon CMG=AD.DB. . To each add LG=CDs, then gnomon CMG + LG = AD.DB + CD', But CMG + LG=BC%; therefore AD.DB+CD'=BC%. Wherefore, if a straight line, &e. 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, or that AC-CD=(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 the rectangle AD.DB, together with the square of CB, is equal to the square of CD. 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 FF. and also through A draw AK parallel to CL or DM. And because AC is equal to CB, the rectangle AL is equal (88.1.) D; |