to the whole, and in like manner situate, having also an Angle common with it, then is that Parallelogram about the fame Diameter with the whole ; which was to be demonstrated. PROPOSITION XXVII. THEORE M. Line, and wanting in Figure by Parallelograms balf Line, being similar to the Defect. LET ET AB be a Right Line, bisected in the Point C, and let the Parallelogram AD be applied to the Right Line AB, wanting in Figure the Parallelogram CE, fimilar and alike situate to that described on half of the Right Line AB. I say, AD is the greatest of all Parallelograms applied to the Right Line A B, wanting in Figure by Parallelograms fimilar and alike fituate to CE. For let the Parallelograin AF be applied to the Right Line AB, wanting in Figure the Parallelogram ÅK, similar and alike fituate to the Parallelogram CE. I say, the Parallelogram AD is greater than the Parallelogram A F. For because the Parallelogram CE is similar to the Parallelogram HK, they stand * about the fame Dia- * 26 of this, meter, let DB their Diameter be drawn, and the Figure described. Then fince the Parallelogram CF is + 43. 1, equal to FE, let HK, which is common, be added; and the whole CH is equal to the whole KE. But CH is f equal to CG, because the Right Line AC is. 36. Į equal to CB. Therefore the whole AF is equal to the Gnomon LNM; and fo CE, that is, the Parallelogram AD is greater than the Parallelogram: AF. Therefore, of all Parallelograms applied to the same Right Line, and wanting in Figure by Parallelograms fimilar and alike fituate, described on the half Line, the greatest is that which is applied to the half Line, being similar to the Defeet; which was to be demon Itrated. PROPOSITION XXVIII. PROBLEM. equal to a Right Line Figure given, deficient by a the Defest of the Parallelegram to be applied. L ET AB be a given Right Line, and let the giv Right-lined Figure, to which the Parallelogram to be applied to the Right Line A B must be equal, be C, which must not be greater than the Parallelogram applied to the half Line, the Defects being similar ; and let D be the Parallelogram, to which the Defect of the Parallelogram to be applied is similar. Now it is required to apply a Parallelogram equal to the given Right-lined Figure C to the given Right Line AB, deficient by a Parallelogram fimilar to D. * 18 of tbàs. Let AB be bisected in E, and on EB describe * the Parallelogram EBFG, similar and alike fituate to D, and complete the Parallelogram AG. Now AG is either equal to C, or greater than it, because of the Determination. If AĞ be equal to C, what was proposed will be done ; for the Parallelogram AG is applied to the Right Line AB, equal to the given Right-lined Figure C, deficient by the Parallelogram EF, similar to the Parallelogram D. But if it be not equal, then HE is greater than C; but EF is equal to HE. Therefore EF fhall also be † 25 of this. greater than C. Now make † the Parallelogram KLMN similar and alike situate to D, and equal to the Excess, by which EF exceeds C. But D is similar to EF: Wherefore KM shall also be similar to EF. Therefore let the Right Line KL be homologous to GE, and LM to GF. Then because EF is equal to C and KM together, EF will be greater than KM; and and so the Right Line GE is greater than KL, and Then fince E F is equal to C and KM together, PROPOSITION XXIX. THE O R E M. equal to a Right-lined Figure given, exceeding ET AB be a given Right Line, and let C be the given Right-lined Figure to which that to be applied to AB must be equal. Likewise let D be the Parallelogram to which the exceeding Parallelogram is to be similar; it is required to apply a Parallelogram to the Right Line AB, equal to the given Rightlined Figure C, exceeding by a Parallelogram similar to D. Bisect AB in E, and let the Parallelogram EL be described * upon the Right Line EB, fimilar and alike * 18. 1. fituate to D, and that + the Parallelogram GH equal † 25 of this, to E L and C together, but similar to D, and alike fituate. Therefore GH is similar to EL; let KH be a Side homologous to FL, and KG to FE, Then becaufe the Parallelogram GH is greater than the Parallelogram EL, the Right Line KH will be greater than FL, and KG greater than FE. Let FL, FE, be produced, and let FLM be equal to KH, FEN equal to KG, and complete the Parallelogram MN. Therefore MN is equal and similar to GÅ; but GH I 1 of bis.is fimilar to EL, and so MN shall be $ fimilar to * 26 of this. EL; and accordingly EL is * about the fame Dia meter with MN. Let: FX be their Diameter, and describe the figure. Then fince GH is equal to EL and C together, as likewise to MN; therefore MN shall be equal to EL and C. Let EL, which is common, be taken away, then the Gnomon Yo* remaining, is equal to C; and fince AE is equal to EB, the Parallelogram AN will be also equal to the Parallelogram EP, that is, to LO; and if EX, which is common, be added, then the whole Parallelogram AX is equal to the Gnomon YO't, but the Gnomon roy is equal to C. Therefore AX shall be also equal to C. Wherefore the Parallelogram AX is applied to the given Right Line AB, equal to the given Right-lined Figure C, and exceeding by the Parallelogram PO, similar to the Parallelogram D; which was to be done. PROPOSITION XXX. PROBLEM. extreme and mean Raiio. , L ET AB be a given terminate Line ; it is required to cut the same according to extreme and mean Ratio. 46. 1. Describe * B C the Square of AB, and apply the - Parallelogram CD to AC, equal to the Square B C, † 29 of this, exceeding + by the Figure AD similar to BC; but BC is a Square, therefore AD shall also be a Square. Now because BC is equal to CD, take away CE which is common; then B F remaining shall be equal to to AD remaining ; but BF is equiangular to AD; therefore the sides that are about the equal Angles are I reciprocally proportional; and so as FE is to I 14 of this. ED, so is AE to'EB, but F E is * equal to AC, that * 34. I. is, to AB, and ED to AE. 'Wherefore as BA is to AE, so is AE to E B, but AB is greater than A E; therefore AE is † greater than EB; and so the Right † 14. 5. Line AB is cut according to extreme and mean Ratio in the Point E; and AE is the greater Segment thereof; which was to be done. Otherwise thus: Let AB be the Right Line given; it is required to cut the same into extreme and mean Ratio. Divide | AB fo in C, that the Rectangle contained i 11. 2. under AB, BC, be equal to the Square of AC. Then because the Rectangle under AB, BC, is equal to the Square of AC, it shall be * as B A is * 17 of this. to AC, fo is AC to CB; and so the Right Li AB is cut into mean and extreme Ratio ; which was to be done. PROPOSITION XXXI. THEOREM. gled Triangle subtending the Right Angle, is e- ET ABC be a rectangular Triangle, having the on BC, is equal to the two Figures together de scribed on. BÅ, AC, which are similar and alike situate to the Figure described on B C. For draw the Perpendicular AD. Then because the Right Line AD is drawn in the Right-angled Triangle ACB, from the Right Angle A, perpendicular to the Base BC; the Triangles ABD, ADC, which are about the Perpendicular AĎ, will be ** 8 of tbisa similar to the whole Triangle ABC, and also to each other. Then because the Triangle ABC is fimilar to the Triangle ABD, it shall be * as CB is to BA, so N 3 is |