PROP. XXVII. THEOR,

Of all parallelograms applied to the same straight line, See N. 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 ČE 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.

+ Hyp,

DLE

#96.6.

F

H43.1.

First, let AK the base of AF, be greater than AC the half of AB: and because CE is similar to the + parallelogram KH, they are about the same * diameter: draw their diameter DB, and complete the scheme: then, because the parallelogram CF is equal to FE, add KH to both; therefore the whole CH is equal to the whole KE: but CH A CKB is equal to CG, because the base AC is

*

*

* 36.1.

†1 Ax.

equal to the base CB; therefore CG is equal to KE: to each of these add CF; then the whole AF is equal† † 2 Ax. to the gnomon CHL: therefore CE or the parallelogram AD, is greater than the parallelogram AF.

*

G FM H

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 the parallelogram DH is equal to the parallelogram DG; wherefore DH is greater than LG: but DH is equal * to DK; therefore DK is greater than LG to each of these add AL; then the whole AD is

M

D

* 34. 1. #36.1.

* 43. 1.

A KC

B

See N.

+27.6.

* 10. 1.

* 18. 6.

† 36.1.

* 25. 6.

Constr. 21.6.

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 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.

*

Τ

X PR

ESB L M

Divide AB into two equal parts* in the point E, and upon EB describe the parallelogram EBFG similar and similarly 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 t to AG; therefore EF also is greater than C. Make* the parallelogram 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 similar to EF: let KL be the homologous side to EG,

*26.6.

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 LK, and GO equal to LM, † 3. 1. and complete the parallelogram XGOP: therefore +31.1. 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: 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 † † 3 Ax. to the remainder C: and because OR is equal to XS, * 43.1. by adding SR to each, the whole OB is equal to the whole XB: but XB is equal to TE, because the base *36. 1. AE is equal to the base EB; wherefore also TE is equal to OB: add XS to each, then the whole TS is + 1 Ax. 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.

PROP. XXIX. PROB.

* 24.6.

To a given straight line to apply a parallelogram equal to See N. 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 C, exceeding by a parallelogram similar to D.

**

18. 6.

25. 6.

Divide AB into two equal parts+ in the point E, † 10.1. and upon EB describe the parallelogram EL similar and similarly situated to D: and make the parallelogram GH equal to EL and C together, and similar and similarly situated to D: wherefore GH is similar* to EL: let KH be the side homologous to FL, and

21.6.

* 26.6.

* 36.1.

* 43. 1.

*24.6.

46. 1. 29.6.

14. 6.

* 34. 1. † 30 Def.

* 14. 5.

K

PX

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 KH, and FEN to KG, and complete
the parallelogram MN: MN is therefore equal and
similar to GH: but GH is si-
milar to EL; wherefore MN
is similar to EL; and conse-
quently EL and MN are about
the same diameter*: 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 AN is equal to the parallelogram NB,
that is, to BM*: 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 figure C, exceeding by the
parallelogram PO, which is similar to D, because PO
is similar to EL. Which was to be done..

*

PROP. XXX. PROB.

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.

Upon AB describe the square BC, and to AC* apply the parallelogram CD, equal to BC, exceeding by the figure AD similar to BC: then, since 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 AD: and these figures are equiangular, therefore their sides about the equal angles are reciprocally proportional: therefore, as FE to ED, so AE to EB: but FE is equal* to AC, 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 AE is greater than EB*: there

*

A

C

F

fore the straight line AB is cut in extreme and mean 3 Def. 6. ratio in E*. Which was to be done.

Otherwise,

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

Divide AB in the point C, so that the rectangle contained by AB, BC may be equal to the square of AC: then, because

A CB

*11. 2.

the rectangle AB, BC, is equal to the square of AC; as BA to AC, so is AC to CB*: therefore AB is cut in * 17. 6. extreme and mean ratio in C*.

Which was to be done. * 3 Def. 6.

PROP. XXXI. THEOR.

In right angled triangles, the rectilineal figure described See N. 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 rectilineal figure described upon BC shall be equal to the similar and similarly described figures upon BA, AC.

Draw the perpendicular+ AD: therefore, because † 12. 1. in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC are similar* to the whole tri- * 8.6. angle ABC, and to one another: and because the triangle ABC is similar to ADB, as CB to BA, so is* * 4.6. BA to BD: and because these three straight lines are proportionals, as the first is to the third, so is the figure upon the first to the similar and similarly described figure upon the second: therefore as CB to BD, so is the figure upon CB to the similar and similarly described figure upon BA: and inversely*, as DB to BC, so is the figure upon BA to that upon BC: for the same reason, as DC to CB, so is the figure upon CA to that upon CB: therefore as BD and DC

B D

* 2 Cor. 20.

6.

* B. 5.

together to BC*, so are the figures upon BA, AC to * 24. 5. that upon BC: but BD and DC together are equal to BC; therefore the figure described on BC is equal* * A. 5. to the similar and similarly described figures on BA, AC. Wherefore, in right angled triangles, &c. Q. E. D.