. tilineal figure kg h is equal (v. 14) to the parallelogram ef: but ef is equal to the figure d; wherefore also kgh is equal to d ; and it is similar to a bc. Therefore the rectilineal figure kgh has been described similar to the figure a b c, and equal to d. Which was to be done. a PROPOSITION XXVI.-THEOREM. If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter. LET the parallelograms a b cd, a efg be similar and similarly situated, and have the angle d a b common : abcd and a efg are about the same diameter. For, if not, let, if possible, the parallelogram bd have its diameter a hc 8 in a different straight line from a f, the diameter of the parallelogram eg: k th and let gf meet a hc in h; and through h draw hk parallel to a d or bc: e therefore the parallelograms a b cd, f akhg being about the same diameter, they are similar to one another (vi. 24): wherefore as d a to a b, so is (vi. def. 1) ga to ak: but because abcd and à efg are similar parallelograms, as b C da is to a b, so is ga to ae; therefore (v. 11) as ga to a e, so is g a to ak; wherefore ga has the same ratio to each of the straight lines a e, ak; and consequently a k is equal (v. 9) to a e, the less to the greater, which is impossible : therefore a bed and akhg are not about the same diameter : wherefore a bcd and a efg must be about the same diameter. Therefore, if two similar parallelograms, &c. Q. E. D. C To understand the three following propositions 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. For example, the parallelogram a c is said to be applied to the straight line a b. a 2. But a parallelogram a e is said to be g applied to a straight line a b, deficient by a parallelogram, when ad the base of a e is less than a b, and therefore a e is less than the parallelogram a c described upon a b in the same angle, and between the same pa a d b f rallels, by the parallelogram dc: and dc is therefore called the defect 3. And a parallelogram ag is said to be applied to a straight line a b, exceeding by a parallelogram, when a f the base of a g is greater than a b, and therefore ag exceeds a c the parallelogram described upon a b in the same angle, and between the same parallels, by the parallelogram bg: of a e. PROPOSITION XXVII.-THEOREM. а ho Of all parallelograms applied to the same straight line, and deficient by pa rallelograms 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 a b be a straight line divided into two equal parts in c, and let the parallelogram ad be applied to the half a c, which is therefore deficient from the parallelogram upon the whole line a b by the parallelogram ce upon the other half cb; of all the parallelograms applied to any other parts of a b, and deficient by parallelograms that are similar and similarly situated to ce, a d is the greatest. Let af be any parallelogram applied to a k, any other part of a b than the half, so as to be deficient from the d 1 e parallelogram upon the whole line a b by the parallelogram kh similar and similarly situated to ce: ad is greater than a f. First, let ak the base of a fbe greater g than ac the half of a b; and because ce is similar to the parallelogram k h, they are about the same diameter (vi. 26); draw their diameter db, and complete the scheme : because the parallelogram cf is equal (i. 43) to fe, add kh to both, therefore the whole ch is equal to the a k whole ke: but ch is equal (i. 36) to cg, because the base a c is equal to the base cb; therefore cg is equal to ke: to each of these add cf; then the whole 8 h af is equal to the gnomon chl, therefore ce, or the parallelogram a d, is greater than the parallelogram a f. Id Next, let a k the base of af be less than a c, and the same construction being made, the parallelogram dh is equal to dg (i. 36), for hm is equal to mg (i. 34), because bc is equal to ca; wherefore dh is greater than lg: but dh equal (i. 43) to dk; therefore dk is greater than 1g; to each of these add al; then the whole a d is greater a k C than the whole af. Therefore, of all b parallelograms applied, &c. Q. E. D. f m PROPOSITION XXVIII.- PROBLEM. 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 defects similar to the defect of that which is to be applied, that is, to the given parallelogram. LET a b 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 applied to the half of the line, having its defect from that upon the whole line similar to the defect of that which is to be applied; and let d be the parallelogram to which this defect is required to be similar. It is required to h 8 apply a parallelogram to the straight line a b, which shall be equal to the figure c, and be deficient from the parallelogram upon the whole line by a parallelogram similar to d. Divide a b into two S a equal parts (i. 10) in the b point e, and upon eb de m scribe the parallelogram e bfg similar (vi. 18) and similarly situated to d, and complete the parallelogram d k n ag, which must either be equal to c, or greater than it, by the determination : and if a gbe equal to c, then what was required is already done : for, upon the straight line a b, the parallelogram a g is applied equal to the figure C, and deficient by the parallelogram ef similar to d: but, if a g be not equal to c, it is greater than it; and ef is equal to ag; therefore ef also is greater than C. Make (vi. 25) the parallelogram klmn equal to the excess of ef above c, and similar and similarly situated to d; but d is similar to ef, therefore (vi. 21) also km is similar to ef. Let kl be the homologous . side to eg, and I m to gf; and because ef is equal to c and k m together, ef is greater than km; therefore the straight line eg is greater than k l, ; and gf than lm: make g x equal to 1k, and go equal to 1 m, and complete the parallelogram xgop: therefore xo is equal and similar to km; but k m is similar to ef; wherefore also xo is similar to ef, and therefore x 0 and e f are about the same diameter (vi. 26): let g p b be their diameter, and complete the scheme. Then, because éf 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. the gnomon ero is equal to the remainder c: and because or is equal (i. 34) to xs, by adding sr to each, the whole ob is equal to the whole Xb: but xb is equal (i. 36) to te, because the base a e is equal to the base e b; wherefore also te 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 a b deficient by the parallelogram sr, similar to the given one d, because sr is similar to ef (vi. 24). Which was to be done. PROPOSITION XXIX.-PROBLEM. To a given straight line to apply a parallelogram equal to a given rectilineal figure, exceeding by a parallelogram similar to another given. LET a b 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 C, exceeding by a parallelogram similar to d. Divide a b into two equal parts in the point e, and upon e b describe (vi. 18) the parallelogram el similar and similarly situated to d: and make (vi. 25) the parallelogram g h equal to el and c together, and similar and similarly situated to d; wherefore gh is similar to el (vi. 21): let kh be the side homologous to fl, and kg to fe: and because the parallelogram gh is greater than el, therefore the side k h is greater than fl, than fe. Produce fl and fe, and make flm equal to k h, and fen to kg, and complete the parallelogram mn: mn is therefore equal and kg х P and similar to gh: but gh is similar to el: wherefore mn is similar to el and consequently el and mn are about the same diameter (vi. 26); draw their diameter fx, and complete the scheme. Therefore since g h is equal to el and c together, and that gh is equal to mn; mn is equal to el and c: take away the common part el; then the remainder, viz. the gnomon nol, is equal to c. And because a e is equal to eb, the parallelogram a n is equal (i. 36) to the parallelogram n b, that is, to bm (i. 43). Add no to each ; therefore the whole, viz. the parallelogram a x, is equal to the gnomon nol. But the gnomon nol is equal to c; therefore also a x is equal to c. Wherefore to the straight line a b there is applied the parallelogram a x equal to the given rectilineal c, exceeding by the parallelogram po, which is similar to d, because po is similar to el (vi. 24). Which was to be done. a PROPOSITION XXX.-PROBLEM. To cut a given straight line in extreme and mean ratio. LET a b be the given straight line, it is required to cut it in extreme and mean ratio. Upon a b describe (i. 46) the square b c, and to a c apply the parallelogram cd, equal to b c, exceeding by the figure ad similar to bc (vi. 29) : but b c is a square, therefore d also a d is a square ; and because bc is equal to cd, by taking the common part ce from each, the remainder bf is equal to the remainder ad; and these figures a ; b are equiangular, therefore their sides about the equal angles are reciprocally proportional (vi. 14): wherefore, as fe to ed, so is ae to eb: but fe is equal to ac (i. 34), that is, to a b; and ed is equal to a e: therefore as ba to a e, so is a e to eb: but a b is greater than ae; wherefore a e is greater than eb (v. 14). There с f fore the straight line a b is cut in extreme and mean ratio in e (vi. def. 3). Which was to be done. Otherwise, Let a b be the given straight line ; it is required to cut it in extreme and mean ratio. Divide a b in the point c, so that the rectangle contained by a b b c be equal to the square of a c (ii. 11): then, because the rectangle a b, bc, is equal to the square of a c; a с b р as b a to a c, so is a c to c b (vi. 17): therefore a b is cut in extreme and mean ratio in c (vi. def. 3). Which was to be done. : زا PROPOSITION XXXI.—THEOREM. In right-angled triangles the rectilineal figure described upon the side oppo site to the right angle is equal to the similar and similarly described figures upon the sides containing the right angle. LET a bc be a right-angled triangle, having the right angle bac. The |