the squares of AC, CB; and because the complement AG is equal to the complement GE, and that AG is* 43. 1. the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also equal to the rectangle AC, CB: therefore AG, GE are equal to twice the rectangle AC, CB: and HF, CK are the squares of AC, CB; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB: but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore if a straight line, &c. Q. E. D. COR. From the demonstration, it is manifest that 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 at the point D; the rectangle AD, DB, together with the square of CD, shall be equal to the square of CB. and also Upon CB describe* the square CEFB, join BE, and * 46. 1. through D draw* DHG parallel to CE or BF; and * 31. 1. through H draw KLM parallel to CB or EF; through A draw AK parallel to CL or BM. cause the complement CH is equal to the comple- *43. 1. ment HF, to each of these equals add DM; therefore the whole CM is equal to the whole DF; but CM is And be * Cor. 4. 2. * K DB C L H M E G F * 36. 1. equal to AL, because AC is equal to CB; therefore also AL is equal to DF: to each of these equals add CH, and the whole AH is equal to DF and CH: but AH is the rectangle contained by AD, DB, for DH is equal* to DB; and DF, together with CH, is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB: to each of these equals add LG, which is equal to the square of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square of CD: but the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB; therefore the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Wherefore, if a straight line, &c. Q. E. D. * Cor. 4. 2. 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. 46. 1. * 31. 1. 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, shall be equal to the square of CD. Upon CD describe the square CEFD, join DE, and through B draw* BHG parallel to CE, or DF: B D H 36. 1. M43. 1. through H draw KLM parallel to AD or EF, and 4. 2. equal to the gnomon CMG: but AM is the rectangle contained by AD, DB, for DM is equal to DB: * Cor. therefore the gnomon CMG is equal to the rectangle AD, DB: add to each of these equals LG, which is equal to the square of CB; therefore the rectangle AD, DB, together with the square of CB, is equal to the gnomon CMG, and the figure LG: but the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c. Q. E. D. PROP. VII. THEOR. 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 shall be equal to twice the rectangle AB, BC, together with the square of AC. Upon AB describe the square ADEB, and con- 46. 1. struct the figure as in the preceding propositions. Because AG is equal to GE, add to each of them CK; * 43. L. therefore the whole AK is equal to the whole CE; *Cor. 4. 2. * 34. 1. E K therefore AK, CE, are double of AK: 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 that 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, shall be equal to the square of the straight line made up of AB and BC together. In AB produced take BD equal to CB, and upon AD describe the square AEFD; and construct two figures such as in the preceding. Because CB is equal to BD, and that CB is equal to GK, and BD to KN; therefore 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 rectangle CK is equal* to BN, * 36. 1. and GR to RN: but CK is equal to RN, because * 43. 1. they are the complements of the parallelogram CO; therefore, also, BN is equal to GR; therefore the four rectangles BN, CK, GR, RN are equal to one another, and so are quadruple of one of them, CK. cause CB is equal to BD, and that BD is equal to BK, that is, to CG; Again, be C B D M G K * Cor. 4. 2. N Cor. P R 4. 2. E H L and CB equal to GK, that is,* to GP; therefore CG is equal to GP; x 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 to PL, because they are the complements of * 43. 1. the parallelogram ML; wherefore AG is also equal to RF: therefore the four rectangles AG, MP, PL, RF are equal to one another, and so are quadruple of one of them, AG. But it was demonstrated, that the four CK, BN, GR, and RN are quadruple of CK: therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK: and because BN is a square,* BK is equal to BD: but BD is equal to BC; * 30 Def. therefore BK is equal to BC; and because AK is the 1. rectangle contained by AB, BC, for BK has been proved equal to BC; therefore four times the rectangle AB, BC is quadruple of AK: but the gnomon AOH was demonstrated to be quadruple of AK; therefore four times the rectangle AB, BC, is equal to the gnomon AOH: to each of these equals add XH, which is equal to the square of AC: therefore four Cor. times the rectangle AB, BC, together with the square of AC, is equal to the gnomon AOH and the square XH: but the gnomon AOH and XH make up the figure AEFD, which is the square of AD: therefore four times the rectangle AB, BC, together with the * 4. 2. |