Wherefore CGKB is a square, and it is upon the fide CB. Book II. For the fame reason HF also is a fquare, and it is upon the fide HG, which is equal to AC; therefore HF, CK are the fquares of AC, CB. And because the complement AG is equal 1 to the complement GE; and because AG=AC.CG h 43. 1. AC.CB, therefore alfo GE=AC.CB, and AG+GE= 2AC.CB. Now, HF AC2, and CK=CB2; therefore, HF + CK + AG + GE➡ AC2 + CB2 + 2AC.CB. h But HF+CK+AG+GE the figure AE, or AB2; therefore AB AC2+CB2+2 AC.CB. Wherefore, if a straight line be divided, &c. Q. E. D. COR. From the demonstration, it is manifeft that the parallelograms about the diameter of a square are likewise fquares. PROP. V. THEOR. F a ftraight line be divided into two equal parts, and alfo into two unequal parts; the rectangle contained by the unequal parts, together with the fquare of the line between the points of fection, is equal to the fquare 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 fquare of CD, is equal to the fquare of CB, or AD.DB+CD2=CB2. Upon CB defcribe a the fquare CEFB, join BE, and a 46. 1. through D draw b DHG parallel to CE or BF; and through b 31. 1. H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM: And because CH-HF, if DM K be added to both, CM DF. But AL CM, therefore AL DF, and adding CH to both, AH A C D M H gnomon CMG. But AHAD.DHAD.DB, because DH DB d; therefore gnomon CMG AD.DB. To each d Cor. 4.24 54 Book II. add LG CD', then gnomon CMG+LG≈ AD.DB+Cd2, But CMG+LG BC, therefore AD.DB + CD2 BC2. Wherefore, if a straight line, &c. Q. E. D. "From this propofition it is manifeft, that'the difference of the fquares of two unequal lines AC, CD, is equal to the rectangle contained by their fum and difference, or that AC2— CD AC+CD. AC-CD." a 46. I. b 31. 1. 36. 1. IF PROP. VI. THE OR. a ftraight line be bifected, and produced to any line thus produced, and the part of it produced, together with the square of half the line bifected, is equal to the fquare of the ftraight line which is made up of the half and the part produced. Let the ftraight line AB be bifected in C, and produced to the point D; the rectangle AD.DB, together with the fquare of CB, is equal to the fquare of CD. Upon CD defcribe a the fquare CEFD, join DE, and through B draw b BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and alfo through A draw AK parallel to CL or DM. And because AC is equal to CB, the rectangle AL C B D C is equal to CH; but A d 43. I. CH is equal to HF; d therefore alfo AL is equal to HF: To each of thefe K add CM; therefore the whole AM is equal to the gnomon CMG. Now AM AD.DM AD.DB, be caufe DM DB. There CB2. fore gnomon CMG=AD.DB, and CMG+LG AD.DB+ But CMG+LG=CF=CD2, therefore AD.DB+CB2 CD. Therefore, if a straight line, &c. Q. E. D. PROP. Book II. IF PRO P. VII. THEOR. Fa ftraight line be divided into any two parts, the fquares 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 ftraight line AB be divided into any two parts in the point C; the fquares of AB, BC are equal to twice the rectangle AB.BC, together with the square of AC, or AB2+ BC2 2AB.BC+AC2. Upon AB defcribe the fquare ADEB, and conftruct the a 46. 1. figure as in the preceding propofitions: Because AG➡GE, AG+CK GE+CK, that is, AK CE, and therefore AK+CE=2AK. But AK+CE gnomon AKF + D 2AB.BC + AC2. Wherefore, if a ftraight line, &c. Q.E. D. Otherwise, "Because AB2=AC2 + BC2 + 2AC.BC2, adding BC2 to 4. 2. both, AB2+BC AC2+2BC2+ 2AC.BC. But BC2+AC.BC= AB.BC b; and therefore, 2 BC2+ 2AC.BC2AB.BC; and therefore AB3 BC3AC2+ 2AB.BC." E 4 "COR. Book II. "COR. Hence, the fum of the fquares of any two lines is "equal to twice the rectangle contained by the lines toge"ther with the fquare of the difference of the lines." PROP. VIII. THEOR. Fa ftraight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the fquare of the other part, is equal to the fquare of the ftraight line which is made up of the whole and the firft-mentioned part. Let the ftraight line AB be divided into any two parts in the point C; four times the rectangle AB.BC, together with the fquare of AC, is equal to the iquare of the straight line made up of AB and BC together. a Produce AB to D, fo that BD be equal to CB, and upon AD defcribe the fquare AEFD; and conftruct two figures fuch as in the preceding. Because GK is equal to CB, and CB to BD, and BD to KN, GK is equal to KN. For the fame 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 43.1. alfo the rectangles GR and RN: But CK is equal to RN, because they are the complements of the parallelogram CO; therefore alfo BN is equal to GR; and the four rect-. angles BN, CK, GR, RN are therefore equal to one another, and fo CK+BN+GR+RN=4CK.. Again, because CB is A d Cor. 4. 2. equal to BD, and BD equald to e 43. 1. MP is equal e to PL, because they b four rectangles AG, MP, PL, RF, are equal to one another, Book II. and fo AG+MP+PL+RF4AG. And it was demonftrated, that CK+BN+CR+RN=4CK; wherefore, adding equals to equals, the whole gnomon AOH=4AK. Now AKAB.BKAB.BC, and 4AK=4AB.BC; therefore, gnomon AOH=4AB.BC; and adding XH, ord AC2, tod Cor. 4. 2. both, gnomon AOH+XH4AB.BC+AC2. But AOH+ XH AF AD2; therefore AD24AB.BC+AC2. Now AD is the line that is made up of AB and BC, added together into one line: Wherefore, if a ftraight line, &c. Q. E. D. "COR. 1. Hence, because AD is the fum, and AC the dif"ference 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 fquare of the fum of the • lines." "COR. 2. From the demonftration it is manifeft, that fince "the fquare of CD is quadruple of the fquare of CB, the "fquare of any line is quadruple of the fquare of half that "line." Otherwife : "Because AD is divided any how in C, AD2=AC2+ a 4. 2. CD2+2CD.AC. But CD 2CB: and therefore CD2= CB2+BD2+2CB.BD a=4CB2, and alfo 2CD.AC-4CB.AC; therefore,AD-AC2+4BC2+4BC.AC. Now BC2+BC.AC= AB.BC; and therefore ADAC2+4AB.BC. Q. E. D." b 3. 2. PROP. |