Wherefore CGKB is a square, and it is upon the side CB. Book II. For the fame reason HF also is a square, and it is upon the fide 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 because AG-AC.CG h 43. r. =AC.CB, therefore also GE=AC.CB, and AG+GE= 2AC.CB. Now, HF=AC2, 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 diameter of a square are likewise squares. I F PROP. V. THE OR. 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+CD2=CB2. Upon CB defcribe a the square CEFB, join BE, and a 46. г. through D drawb 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 с 36. 1. therefore AL=DF, and adding CH to both, AH= gnomon CMG. But AH = AD.DH=AD.DB, because DH DBd; therefore gnomon CMG AD.DB. To each d Cor. 4.2+ Book II. add LG=CD2, then gnomon CMG+LG=AD.DB+CD, But CMG+LG = BC2, therefore AD.DB + CD2 =BC2. 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, or that AC2CD AC+CD. AC-CD." a 46. 1 b 31. 1. I F PROP. VI. THEOR. 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 a the square 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 also through A draw AK parallel to CL or DM. And because AC is equal fore 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. Book II. IF PROP. VII. THEOR. F 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+ BC22AB.BC+AC2. Upon AB describe the square ADEB, and construct the a 46. 1. figure as in the preceding propositions: Because AG=GE, AG+CK=GE+CK, that is, AK= CE, and therefore AK+CE-2AK. A But AK+CE = gnomon AKF + C B 2AB.BC + AC2. Wherefore, if a straight line, &c. Q. E. D. Otherwise, "Because AB=AC2 + BC2 + 2AC.BC, adding BC2 to 4. 2. both, AB2+BC2=AC2+2BC2+ 2AC.BC. But BC2+AC.BC= AB.BC b; and therefore, 2BC2+ 2AB.BC." 2AC.BC2AB.BC; and therefore AB2 + BC2 = AC2 + b 3. 2: Book II. "Cor. Hence, the fum of the squares of any two lines is " equal to twice the rectangle contained by the lines toge "ther with the square of the difference of the lines." 1 F PROP. VIII. THEOR. 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. 43. 1. 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 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 GK to KN, the rectangles CK and BN are equal, as also 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 rectangles BN, CK, GR, RN are therefore equal to one another, and fo CK+BN+GR+RN=4CK. Again, because CB is d Cor. 4. 2, equal to BD, and BD equald to A € 43. 1. BK, that is, to CG; and CB equal four four rectangles AG, MP, PL, RF, are equal to one another, Book II. and fo AG+MP+PL+RF=4AG. And it was demonstrated, that CK+BN+CR+RN=4CK; wherefore, adding equals to equals, the whole gnomon AOH=4AK. Now AKAB.BK=AB.BC, and 4AK-4AB.BC; therefore, gnomon AOH=4AB.BC; and adding XH, ord AC2, tod Cor. 4. 2. both, gnomon AOH+XH=4AB.BC+AC2. But AOH+ XH=AF=AD2; therefore AD4AB.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 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 square of the fum of the ' lines." "COR. 2. From the demonstration it is manifest, that fince "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." Otherwife: i " Because AD is divided any how in Ca, ADAC2 + a 4.2. 2 CD2+2CD.AC. But CD = 2CB: and therefore CD2= PROP. |