the angle ABC is equal to AEF, and BAC common to the two triangles BAC, EAF; therefore (VI. 4. cor.) as AB: BC:: AE EF. And, (V. 7.) because the opposite sides of parallelograms are equal to one another, AB: AD::AE: AG; and DC: BC:: GF : EF; and also CD: DA::FG:GA. Therefore the sides of the parallelograms BD, EG about the equal angles are proportionals: the parallelograms are, therefore, (VI. def. 1.) similar to one another. In the same manner, it would be shown that the parallelogram BD is similar to HK. Therefore each of the parallelograms EG, HK is similar to BD. But (VI. 21.) rectilineal figures which are similar to the same figure, are similar to one another; therefore the parallelogram EG is similar to HK.* Wherefore the parallelograms, &c.

PROP. XXV. PROB.

To describe a rectilineal figure which will be similar to one given rectilineal figure, and equal to another. Let ABC and D be given rectilineal figures. It is required to describe a figure similar to ABC, and equal to D.

Upon the straight line BC describe (I. 45. cor.) the parallelogram BE equal to ABC; also upon CE describe the parallelogram CM equal to D, and having the angle FCE equal to CBL. Therefore (I. 29. and 14.) BC and CF are in a straight line, as also LE and EM. Between BC and CF find (VI. 13.) a mean proportional GH, and on it describe (VI. 18.) the figure GHK similar, and similarly situated, to ABC: GHK is the figure required. Because BC: GH :: GH: CF, and if three straight lines be proportionals, as the first is to the third, so is (VI. 20. cor. 2.) the figure upon the first to the similar and similarly described figure upon the second; therefore

D

AA

B

L

F

as BC to CF, so is ABC to KGH: but (VI. 1.)
as BC to CF, so is BE to EF:

H

therefore (V. 11.) as ABC is to KGH, so is BE to EF. But (const.) ABC is equal to BE; therefore KGH is equal (V. 14.) to EF and (const.) EF is equal to D; wherefore also KGH is equal to D; and it is similar to ABC. Therefore the rectilineal figure KGH has been described, similar to ABC, and equal to D: which was to be done.

* Hence, GF: FE:: FH: FK. Therefore the sides of the parallelograms GK and EH, about the equal angles at F, are reciprocally proportional; and (VI. 14.) these parallelograms are equal; a conclusion which agrees with the 43d proposition of the first book.

M

PROP. XXVI. THEOR.

If two similar parallelograms have a common angle, and be similarly situated; they are about the same diagonal.

Let the parallelograms BD, EG be similar and similarly situated, and have the angle DAB common: they are about the same diagonal.

For, if not, let, if possible, the parallelogram BD have its diagonal AHC in a different straight line from AF, the diagonal of EG, and let GF meet AHC in H; and through H draw HK parallel to AD or BC. Therefore (VI. 24.) since the parallelograms BD, KG are about the same diagonal, they are similar to one another wherefore (VI. def. 1.)

DA: AB:: GA: AK.

But (VI. def. 1.) because BD, and EG are similar,

DA: AB:: GA: AE;

A

G

D

K

H

E

F

B

therefore (V. 11.) as GA: AE :: GA: AK; wherefore (V. 9.) AK is equal to AE, the less to the greater, which is impossible. Therefore AHC is not the diagonal of BD; and in the same manner it may be shown that no other line, except AFC, is the diagonal of BD: AFC therefore is its diagonal, and BD, EG are about the same diagonal AFC. Therefore, if two similar parallelograms, &c.

PROP. XXVII. THEOR.

Or all the parallelograms that can be inscribed in any triangle, that which is described on the half of one of the sides as base, is the greatest.

Let ABC be a triangle, having BC one of its sides bisected in D; draw (I. 31.) DE parallel to BA, and EF to BC: let also G be any other point in BC, and describe the parallelogram GK: FD is greater than KG.

K

B

F

E M

T.

N

H

D G

c

If G be in DC, through C draw CL parallel to BA, and produce FE, KH, GH, as in the figure. Then (I. 43.) the complements LH and HD are equal; and since the bases CD, DB are equal, the parallelograms ND, DK (I. 36.) are equal. To LH add ND, and to HD add DK: then (I. ax. 2.) the gnomon MND is equal to the parallelogram KG. But (I. ax. 9.) DL is greater than MND; and therefore FD, which (I. 36.) is equal to DL, is greater than KG, which is equal to the gnomon MND.

If G were in BD, since BD is equal to DC, AE is equal (VI. 2.) to EC, and AF to FB; and by drawing through A a parallel to BC, meeting DE produced, it would be proved in the same manner, that FD is greater than the inscribed parallelogram applied to BG. Therefore, of all the parallelograms, &c. *

Cor. Since (VI. 24.) all parallelograms having one angle coinciding with BCL, and their diagonals with CA, are similar, it follows from this proposition, that if, on the segments of a given straight line BC, two parallelograms of the same altitude be described, one of them DL, similar to a given parallelogram, the other DF will be the greatest possible, when the segments of the line are equal.

PROP. XXVIII. PROB.

To divide a given straight line into two parts, such that parallelograms of equal altitude may be described upon them, one equal to a given rectilineal figure, and the other similar to a given parallelogram; the rectilineal figure not being greater (VI. 27.) than the parallelogram described on half the given line, and similar to the given parallelogram.

L

H R G

V

Let AB be a given straight line, and CD a given parallelogram, of which EC is a side; let AB be bisected (I. 10.) in F; and on FB (VI. 18.) let the parallelogram FG be described, similar to CD, and having FB homologous to EC; let also K be a rectilineal figure, which is not greater than FG: it is required to divide AB into two parts, such, that if on one of them a parallelogram be described, similar to CD, and having that part homologous to EC, the parallelogram on the other, of equal altitude, may be equal to K.

K

0 M

D

E

Through A draw (I. 31.) AL parallel to FH, and produce GH

* The parallelogram FD exceeds KG by the parallelogram OM similar to DL or DF, and described on OH, which is equal to DG, the difference of the bases BD and BG. Hence we derive the solution of the next proposition.

The enunciation of this proposition here given is much more simple and intelligible than that of Euclid, and the proof is considerably shortened. Euclid's enunciation, as given by Dr. Simson, is as follows: "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." It may be remarked, that this proposition, in its simplest case, is the same as the second corollary to the fifth proposition of the second book.

Respecting this proposition and the next, see the Notes at the end of the

book.

to meet it in L. Then, if AH be equal to K, the thing is done : for on FB one of the parts of AB, the parallelogram FG is described similar to CD, and similarly situated; and on AF, the remaining part, a parallelogram AH, of equal altitude, is described, equal to the given rectilineal figure K.

L

H R G

K

A

N

V

E C

But if AH be not equal to K, it is greater by the determination therefore also FG, which (I. 36.) is equal to AH, is greater than K. Make (VI. 25.) the parallelogram MN similar to CD or FG, and equal to the excess of FG above K. Let the side NO be homologous to HF or DE, and consequently NP to HG: and since FG is equal to K and MN taken together, it is plain that HF, HG are respectively greater than NO, NP. Cut off, therefore, HQ, HR respectively equal to NO, NP, and complete the parallelogram HQSR; produce also the sides QS and RS. Then, since OP is similar to FG, QR, which is equal and similar to OP, is also (VI. 21.) similar to FG, and therefore (VI. 26.) QR and FG are about the same diagonal. Let HSB be their common diagonal. Now, since (const.) K and OP are together equal to FG, take away OP and QR, which are equal, and (I. ax. 3.) there remains the figure K equal to the gnomon RVF. But it may be shown as in the last proposition, that TX is equal to the gnomon RVF: therefore TX is equal to K; and AB is divided into two parts AX, XB, such that on one of them the parallelogram TX is described equal to K, and on the other the parallelogram XV, of the same altitude, is described similar to the given parallelogram CD, and having the segment XB, of the given line, homologous to EC: which was to be done.

PROP. XXIX. PROB.

O M

To produce a given straight line so that a parallelogram similar to a given one being described on the produced part, another parallelogram of equal altitude described on the whole line produced, may be equal to a given rectilineal figure.

Let AB be a given straight line, C a given rectilineal figure, and D a given parallelogram: it is required to produce AB, so that if a parallelogram be described on the produced part similar to D, and similarly situated, a parallelogram of the same altitude on the whole produced line shall be equal to C.

Bisect AB in E, and (VI. 18.) on EB describe the parallelogram EL similar, and similarly situated to D: and make (VI. 25.) the parallelogram KH equal to EL and C together, and similar,

and similarly situated to D; wherefore (VI. 21. and const.) KH is similar to EL. Let GH be the side homologous to FL, and GK to FE: and because KH is greater than EL, the side GH is greater than FL, and GK than FE. Produce FL and FE,

and make FLM equal to GH, and FEN to GK, and complete the parallelogram MN. MN is therefore equal and similar to KH; but KH is similar to EL; wherefore (VI. 21.) MN is similar to EL, and consequently (VI.

:

D

C

L M

E

B

N

H

26.) EL and MN are about the same diagonal draw their diagonal FBX, and complete the figure. Therefore, since KH is equal to EL and C together, and that KH is equal to MN; MN is equal to EL and C: take away the common part EL; then the remaining gnomon NOL is equal to C. And because AE is equal to EB, the parallelogram AN is equal (I. 36.) to NB, that is, (I. 43.) to BM. Add NO to each; therefore the whole parallelogram AX, is equal to the gnomon NOL. But NOL is equal to C; therefore also AX is equal to C. Wherefore, to the whole produced line AO there is applied the parallelogram AX equal to the given figure C; and the parallelogram PO, of the same altitude as AX, is described on the produced part BO, and is similar to D, because (VI. 24.) PO is similar to EL: which was to be done.

PROP. XXX. PROB.

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

Let AB be the given straight line; it is required to divide it in extreme and mean ratio.

C

B

Divide AB (II. 11.) in the point C, so that the rectangle AB.BC may be equal to the square of AC: then, (VI. 17.) because the rectangle AB.BC is equal to à the square of AC, as BA: AC:: AC: CB: therefore (VI. def. 3.) AB is cut in extreme and mean ratio in C: which was to be done.

THE rectilineal figure described upon the hypotenuse of a right-angled triangle, is equal to the similar, and similarly described figures upon its other sides.

Let ABC be a right-angled triangle, having the right angle BAC: the rectilineal figure E, described upon BC, is equal to F, G, the similar figures, similarly described upon BA, AC.