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they are about the same * diameter: draw their diameter *26.6. DB, and complete the scheme: then, because

DLE

43. L.

F

G

H

66. 1.

the parallelogram CF is equal* to FE, add
KH to both; therefore the whole CH is equal
to the whole KE: but CH is equal 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 AD, is
greater than the parallelogram AF.

A CKB

+1 Ax.

+2 Ax.

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 greater than the whole AF. Therefore, of all parallelograms applied, &c. Q. E. D.

CFM H

LAD

E

34. 1. 36.1.

AKC

B

43.1.

PROPOSITION XXVIII.

PROB-To a given straight line to apply a parallelogram See N. 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 +27.6. 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.

Divide AB into two equal parts * in the point E, and upon 10.1.

18.6.

+36.1.

25.6.

+ Constr. 21.6.

† 3.1.

† 31.1.

26.6.

+3 Ax. 43.1. *36.1.

+ 1 Ax.

24.6.

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T

X

PR

A

ESB

L

M

C

D

K

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 determi-
nation. If AG be equal to C, then what was required is al-
ready 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 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 similart to EF, therefore *
also KM is similar to EF: let KL be the homologous side to
EG, 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, and complete + the parallelo-
gram XGOP; therefore 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 equalt to the remainder C: and
because OR is equal * to XS, by adding SR to each, the whole
OB is equal to the whole XB: but XB is equal * to TE, be-
cause 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 there-
fore 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, si-
milar to the given one D, because SR is similar * to EF.
Which was to be done.

See N.

PROPOSITION XXIX.

PROB. 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 AB which shall be equal to the figure C, exceeding by a parallelogram similar to D.

K

H

Divide AB into two equal parts † in the point E; and † 10. 1. upon EB, describe * the parallelogram EL similar and simi- * 18.6. larly 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 KG to FE: and because the parallelogram GH is greater than EL, therefore the side KH is greater than FL, and KG

25.6.

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F LM

* 21.6.

D

E

BO

N

PX

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 similar to EL; wherefore MN is similar to EL; and consequently EL and MN are about the same diameter*: draw their diameter FX, * 26.6. 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 * 24.6. to be done.

36. 1. 43. 1.

:

1

M

46.1. 29.6.

PROPOSITION 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.

A

D

EB

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

* 14.6. 34. 1.

+ 30 Def.

*14.5.

* 11.2.

*17.6.

be done.

Otherwise:

C F

Let AB be the given straight line; it is required to cut 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 the rectangle AB, BC is equal to A CB the square of AC, as BA to AC, so is AC to CB *:

*3 Def. 6. therefore AB is cut in extreme and mean ratio in C*. Which

See N.

was to be done.

PROPOSITION XXXI.

THEOR. 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 rectilineal figure described upon BC, shall be equal to the similar and similarly described figures upon BA, AC.

Draw the perpendicular + AD: therefore, because in the † 12. 1. right-angled triangle ABC, AD is drawn from the right angle

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6.

B. 5.

is the figure upon the first to the similar and similarly described figure* upon the second: therefore as CB to BD, so is 2 Cor. 20. 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 together to BC *, so are the figures upon BA, AC, to that upon BC: but BD and DC together are equal to BC; therefore the figure described on BC is equal to the similar and similarly described figures upon BA, AC. Wherefore, in right-angled triangles, &c. Q. E. D.

PROPOSITION XXXII.

24.5.

A. 5.

THEOR. If two triangles, which have two sides of the one, See N. proportional to two sides of the other, be joined at one angle so as to have their homologous sides parallel to one another, the remaining sides shall be in a straight line.

Let ABC, DCE be two triangles, which have the two sides BA, AC proportional to the two CD, DE, viz. BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE: BC and CE shall be in a straight line.

Because AB is parallel to DC, and the straight line AC meets them, the alternate angles * BAC, ACD are equal; for the 29. 1.

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the triangle ABC is equiangular* to DCE;

6.6.

therefore the angle ABC is equal to the angle DCE: and the angle BAC was proved to be equal to ACD; therefore the whole

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