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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 liné AB, which shall be equal to the figure H G OF C, and be deficient from the parallelogram upon the whole line by a parallelogram similar

X to D.

IP

T
Divide AB in two equal

R
parts (10. 1.) in the point E,
and upon EB describe the pa A E SB
rallelogram EBFG similar (18.
6.) and similarly situated to

L

M
D, and complete the parallelo-

С
gram AG, which must either
be equal to C or greater than

K

N it, by the determination: and 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 to AG; therefore EF also is greater than C. Make (25. 6.) 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 (21. 6.) also KM is similar to EF; let KL

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. See Note:

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 parallelogram 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 (26. 6.): 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 (34. 1.) to XS, by adding SR to each, the whole OB is equal to the whole XB: but XB is equal (36. 1.) 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 AB deficient by the parallelogram SR, similar to the given one D, because SR is similar to EF (24. 6.). Which was to be done.

PROP. 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

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 AB into two equal parts in the point E, and upon EB describe (18. 6.) the parallelogram EL similar and similarly si

similar.

See Note.

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L

C

F

D

B

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

tuated to D: and make (25. 6.) the parallelogram GH equal to 1 db
EL and C together, and similar and similarly situated to D; :
wherefore GH is similar to EL (21. 6.); let KH be the side
homologous to FL, and KG to FE; and because the parallelo-
gram 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

K

H
equal and similar to
GH, but GH is simi-
lar to EL; wherefore
MN is similar to EL,
and consequently EL
and MN are about the

M
same diameter (26. 6.):
draw their diameter
FX, and complete the
scheme. Therefore,

E
since GH is equal to
EL and C together, and
that GH is equal to

N
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
(36. 1.) to the parallelogram NB, that is, to BM (43. 1.). Add
NO to each; therefore the whole, viz. the parallelogram
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 given rectilineal C, exceeding by the parallelogram PO, which is similar to D, because PO is similar to EL (24. 6.) Which was to be done.

PROP. XXX. PROB. "extreme and mean ratio.

P X

AX is

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Upon AB describe (46. 1.) the square BC, and to AC apply the parallelogram CD equal to BC, exceeding by the figure AD similar to BC (29. 6.): 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 remain- A

B der BF is equal to the remainder AD: and these figures are equiangular, therefore their sides about the equal angles are reciprocally proportional (14.6.); wherefore as FE to ED, so AE to EB; but FE is equal to AC (34. 1.), that is, to AB; and ĖD is

С F equal to AĖ: therefore as BA to AE, so is AE to EB: but AB is greater than AE; wherefore AE is greater than EB (14. 5.): therefore the straight line AB is cut in extreme and mean ratio in E (3. def. 6.). Which was to be done.

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Otherwise.
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 be equal to the square of AC (11. 2.).
Then, because the rectangle AB, BC is equal to A С B

square of AC, as BA to AC, so is AC to CB (17. 6.): therefore AB is cut in extreme and mean ratio in C (3. def. 6.).

was to be done.

the

Which

PROP. XXXI. THEOR. Is 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 is equal to the similar and similarly described figures upon BA, AC.

Draw the perpendicular AD; therefore, because 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 triangle ABC, and to one another (8. 6.), and

* See Note.

because the triangle ABC is similar to ADB, as CB to BA, so
is BA to BD (4. 6.); and because these three straight lines are
proportionals, as the first to the third, so is the figure upon the
first to the similar and similarly described figure upon the se
cond (2 Cor.): therefore as CB to

A
BD, so is the figure upon CD to the
similar and similarly described fi-
gure upon BA: and, inversely (B.
5.), as DB to BC, so is the figure
upon BA to that upon BC; for the
same reason, as DC to CB, so is B

D
the figure upon CA to that upon
CB. Wherefore as BD and DC
together to BC, so are the figures upon BA, AC to that upon
BC (24. 5.): but BD and DC together are equal to BC. There-
fore 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.

PROP. XXXII. THEOR.

If two triangles which have two sides of the one 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.*

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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 he parallel to DC, and AC to DE. BO and CE are in a straight line.

Because AB is parallel to DC, A
and the straight line AC meets
them, the alternate angles BAC,

D
ACD are equal (29. 1.); for the
same reason, the angle CDE is
equal to the angle ACD; where-
fore also BAC is equal to CDE;
and because the triangles ABC,

E

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TY L

See Note.

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