PROP. XXVI. THEOR. If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter. Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common: ABCD and AEFG shall be about the same diameter. K For, if not, let, if possible, G D H lel to AD or BC: therefore the parallelograms ABCD, AKHG being about the same diameter, *24.6. they are similar to one another; wherefore as *1 Def.6. DA to AB, so is* GA to AK: but because + Hyp. ABCD and AEFG are similar+ parallelograms, *11.5. as DA is to AB, so is GA to AE; therefore* as *9.5. GA to AE, so GA to AK; that is, GA has the same ratio to each of the straight lines AE, AK; and consequently AK is equal to AE, the less to the greater, which is impossible: therefore ABCD and AKHG are not about the same diameter wherefore ABCD and AEFG must be about the same diameter. Therefore, if two similar, &c. Q. E. D. To understand the three following proposi'tions more easily, it is to be observed, 1. That a parallelogram is said to be applied to a straight line, when it is described upon it as one of its sides. Ex. gr. the parallelogram 'AC is said to be applied to the straight line 'AB. 2. But a parallelogram AE is said to be ap'plied to a straight line AB, deficient by a parallelogram, when AD the base of AE is less than AB, and therefore AE is less than the parallelogram AC described upon AB in the same angle, and between the same parallels, by the parallelogram DC; and DC is therefore called the defect of AE. 3. And a parallelogram AG is said to be applied to a straight line AB, exceeding by a parallelogram, when AF the base of AG is greater than AB, and therefore AG exceeds AC the paralle.gram described upon AB in the same angle, and between the same parallels, by the 'parallelogram BG.' PROP. XXVII. THEOR. Of all parallelograms applied to the same straight line, and deficient by parallelograms, similar and similarly situated to that which is described upon the half of the line; that which is applied to the half, and is similar to its defect, is the greatest. Let AB be a straight line divided into two equal parts in C; and let the parallelogram AD be applied to the half AC, which is therefore deficient from the parallelogram upon the whole line AB by the parallelogram CE upon the other half CB: of all the parallelograms applied to any other parts of AB, and deficient by parallelograms that are similar and similarly situated to CE, AD shall be the greatest. Let AF be any parallelogram applied to AK, any other part of AB than the half, so as to be deficient from the parallelogram upon the whole line AB by the parallelogram KH similar and similarly situated to CE: AD shall be greater than AF. First, let AK the base of AF be greater than AC the half of AB: and because CE is similar† † Hyp. to the parallelogram KH, they are about the *26.6. same * diameter: draw their diameter DB, and complete the scheme: then, because the paral*43.1. lelogram CF is equal to FE, add KH to both; therefore the whole CH is equal to the whole *36. 1. KE: but CH is equal to CG, A because the base AC is equal to * +1 Ax. the base CB; therefore CG is equal to KE: to each of these add CF; then the whole AF is +2 Ax. equal to the gnomon CHL: therefore CE, or the parallelogram AD is greater than the parallelogram AF. * Next, let AK the base of AF be less than AC: then, the same construction being made, because BC is equal to CA, therefore HM is equal to MG; therefore *36. 1. the parallelogram DH is equal* to the parallelogram DG; wherefore DH is greater than LG : *43. 1. but DH is equal* to DK; there *S4. 1. † 27.6. fore DK is greater than LG: to each of these add AL; then the whole AD is greater than the whole AF. Therefore, of all parallelograms applied, &c. Q. E. D. PROP. XXVIII. PROB. To a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by a parallelogram similar to a given parallelogram: but the given rectilineal figure to which the parallelogram to be applied is to be equal, must not be greater than the parallelogram applied to half of the given line, having its defect similar to the defect of that which is to be applied; that is, to the given parallelogram. Let AB be the given straight line, and C the given rectilineal figure, to which the parallelogram to be applied is required to be equal, which figure must not be greater than the parallelogram Divide AB into two equal parts* in the point E, and upon EB describe H G OF T R A E SB L M C K *10. 1. the parallelogram EBFG similar* and similarly * 18. 6. situated to D, and complete the parallelogram AG, which must either be equal to C, or greater than it, by the determination. If AG be equal to C, then what was required is already done : for, upon the straight line AB the parallelogram AG is applied equal to the figure C, and deficient by the parallelogram EF similar to D. But, if AG be not equal to C, it is greater than it and EF is equal to AG; therefore EF +36. 1. also is greater than C. Make the parallelogram *25.6. KLMN equal to the excess of EF above C, and similar and similarly situated to D: then, since D is similar to EF, therefore also KM is si- † Const. milar to EF: let KL be the homologous side to EG, and LM to GF: and because EF is equal to C and KM together, EF is greater than KM; therefore the straight line EG is greater than KL, and GF than LM: make+ GX equal to † 3.1. LK, and GO equal to LM, and complete+ the † 31. 1. parallelogram XGOP: therefore XO is equal and similar to KM: but KM is similar to EF; wherefore also XO is similar to EF; and therefore* XO and EF are about the same diameter: * 26.6. let GPB be their diameter, and complete the *21. 6. scheme. Then, because EF is equal to C and KM together, and XO a part of the one is equal to KM a part of the other, the remainder, viz. +3 Ax. the gnomon ERO, is equal to the remainder 43. 1. C: and because OR is equal to XS, by adding SR to each, the whole OB is equal to the whole *36.1. XB: but XB is equal to TE, because the base * AE is equal to the base EB; wherefore also TE +1 Ax. is equal to OB: add XS to each, then the whole TS is equal to the whole, viz. to the gnomon ERO: but it has been proved that the gnomon ERO is equal to C; and therefore also TS is equal to C. Wherefore the parallelogram TS, equal to the given rectilineal figure C, is applied to the given straight line AB deficient by the parallelogram SR, similar to the given one D, because SR is similar* to EF. Which was to be done. 24.6. 10. 1. PROP. XXIX. PROB. To a gwen straight line to apply a parallelogram equal to a given rectilineal figure, exceeding by a parallelogram similar to another given. Let AB be the given straight line, and C the given rectilineal figure to which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given line is required to be similar. It is required to apply a parallelogram to the given straight line AB which shall be equal to the figure Č, exceeding by a parallelogram similar to D. Divide AB into two equal parts + in the point *18. 6. E, and upon EB describe the parallelogram EL *25.6, similar and similarly situated to D: and make* the parallelogram GH equal to EL and C together, and similar and similarly situated to D: 21.6. wherefore GH is similar to EL: let KH be the side homologous to FL, and KG to FE: and |