tilineal figure k g h is equal (v. 14) to the parallelogram ef: but ef is equal to the figure d; wherefore also k g h is equal to d; and it is similar to abc. Therefore the rectilineal figure k g h has been described similar to the figure a bc, and equal to d. Which was to be done.

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 dab common: abcd and aefg are about the same diameter.

For, if not, let, if possible, the parallelogram b d have its diameter ahc in a different straight line from af, the diameter of the parallelogram eg, and let gf meet ahcin h; and through h draw hk parallel to ad or bc: e therefore the parallelograms abcd, akh g being about the same diameter, they are similar to one another (vi. 24): wherefore as da to a b, so is (vi. def. 1) ga to ak but because abcd and aefg are similar parallelograms, as b

:

a

c

d

da is to a b, so is ga to ae; therefore (v. 11) as ga to ae, so is ga 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 abcd and akh g are not about the same diameter: wherefore a bed and a efg must be about the same diameter. Therefore, if two similar parallelograms, &c. Q. E. D.

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 ac is said to be applied to the straight line a b.

c

2. But a parallelogram a e is said to be applied to a straight line ab, deficient by a parallelogram, when ad the base of ae is less than a b, and therefore ae is less than the parallelogram ac described upon ab in the same angle, and between the same pad b rallels, by the parallelogram dc: and dc is therefore called the defect of a e.

a

f

3. And a parallelogram a g is said to be applied to a straight line a b, exceeding by a parallelogram, when af the base of a g is greater than a b, and therefore ag exceeds a c the parallelogram described upon ab in the same angle, and between the same parallels, by the parallelogram bg.

PROPOSITION XXVII.-THEOREM.

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 a c, 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 a b, and deficient by parallelograms that are similar and similarly situated to ce, ad is the greatest.

d 1

e

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 parallelogram upon the whole line ab 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 than ac the half of a b; and because ce is similar to the parallelogram kh, 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 whole ke but ch is equal (i. 36) to cg,

:

because the base ac is equal to the base cb; therefore cg is equal to ke:

to each of these add cf; then the whole af is equal to the gnomon chl, therefore ce, or the parallelogram a d, is greater than the parallelogram a f.

Next, let ak the base of af be less than ac, 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 1g: but dh is equal (i. 43) to dk; therefore dk is greater than 1g; 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.

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.

h

g o f

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 apply a parallelogram to the straight line ab, which shall be equal to the figure C, and be deficient from the parallelogram upon the whole line by a parallelogram similar to d.

k

n

a g, which must either be equal to c, or greater than it, by the determination: and if a g be equal to c, then what was required is already done: for, upon the straight line ab, 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 a g; 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 kl, and gf than 1m: make gx equal to 1k, and go equal to 1m, and complete the parallelogram xgop: therefore xo is equal and similar to km; but km is similar to ef; wherefore also xo is similar to ef, and thereforexo and e f are about the same diameter (vi. 26): let gpb be their diameter, and complete the 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. 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

ae is equal to the base eb; 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 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 a b, 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 eb describe (vi. 18) the parallelogram el similar and similarly situated to d: and make (vi. 25) the parallelogram gh 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 kh is greater than fl, and kg 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 similar to gh: but gh is similar to el: wherefore m n 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 gh 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 ae 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 ax, is equal to the gnomon nol. But the gnomon nol is equal to c; therefore also ax is equal to c. Wherefore to the straight line ab there is applied the parallelogram ax 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.

PROPOSITION XXX.-PROBLEM.

To cut a given straight line in extreme and mean ratio.

LET ab be the given straight line, it is required to cut it in extreme and mean ratio.

d

Upon ab describe (i. 46) the square b c, and to a c apply the parallelogram cd, equal to bc, exceeding by the figure ad similar to be (vi. 29): but bc is a square, therefore also ad 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 a d; and these figures a 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 ab; and ed is equal to ae: therefore as ba to ae, so is ae to eb: but ab is greater than ae; wherefore a e is greater than eb (v. 14). Therefore the straight line a b is cut in extreme and mean ratio in e (vi. def. 3). Which was to be done. Otherwise,

C

f

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, bc be equal to the square of a c (ii. 11): then, because the rectangle ab, bc, is equal to the square of ac; as ba to a c, so is a c to cb (vi. 17): therefore a b is cut in extreme and mean ratio in c (vi. def. 3). Which was to be done.

a

c b

PROPOSITION XXXI.-THEOREM.

In right-angled triangles the rectilineal figure described upon the side opposite to the right angle is equal to the similar and similarly described figures upon the sides containing the right angle.

LET abc be a right-angled triangle, having the right angle bac. The