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Book VI. ted to D: And make b the parallelogram GH equal to EL and

m C together, and similar and similarly situated to D; whereb 25. 6.

fore GH is similar to EL : Let KH be the fide homologous to c 21, 6.

FL, and KG to FE: And because the parallelogram GH is
greater than EL, therefore the fide 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 e-
qual and similar to K K
GH; but GH is fi-
milar to EL; where
fore MN is similar to
EL, and consequent-

G
ly EL and MN are
about the same dia-

L M

F
d 20..8
meter d: Draw their

D
diameter FX, and
complete the scheme.

E

B

0
Therefore since GH
is equal to EL and C
together, and that

N

Р Х
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 e 36. I. because AE is equal to EB, the parallelogram AN is equal to f 43. s.

the parallelogram NR, that is, to BMf. 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 limilar to EL, Which was to be done.

8 24. 6.

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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

Upon AB describe a the square BC, and to AC apply the Book VI. parallelogram CD equal to BC, exceeding by the figure AD fi- m milar to BC : But BC is a square,

a 46. I. therefore also AD is a square; and be

b 29. 6.

D cause BC is equal to CD, by taking the common part CE from each, the remainder BF is equal to the remainder

А

E B AD: And these figures are equiangu"lar, therefore their lides about the equal

C 14. 6. angles are reciprocally proportionale: Wherefore as FE to ED, fo AE to EB :

d 34. I. But FE is equal to AC4, that is, to AB; and ED is equal to AE: Therefore as BA to AE, so is AE to EB: But AB is

C С greater than AE; wherefore AE is greater than EB 4: Therefore the straight line AB is cut in ex

f def. 6. treme and mean ratio in E f. which was to be done.

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 AC3.

8 II, Then, because the rectangle AB, BC is equal A

CB to the square of AC, as BA to AC, so is AC to CB h: Therefore AB is cut in extreme and mean ratio h 17. 6. in Cf, Which was to be done.

e 14. 5.

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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 is equal to the fimilar, 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 limilar to the whole triangle ABC, and to one another“, a 8. 6.

and

20. 6.

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Book VI. and because the triangle ABC is similar to ABD, as CB to

W BA, fo is BA to BD, and because these three straight lines b 4. 6.

are proportionals, as the first to the third, so is the figure up

on the first to the similar, and similarly described figure upon C 2. Cor.

the second: Therefore as CB
to BD, so is the figure upon
CB to the similar and fimi-
larly described figure upon
BA: And, inverselyd, as DB
to BC, so is the figure upon
BA to that upon BC: For
the same reason, as DC to

B

D
CB, fo is the figure upon CA
upon

CB. Wherefore
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 f to the similar
and fimilarly described figures on BA, AC. Wherefore, in
right angled triangles, &c. Q. E. D.

С

to that

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f A. S.

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See N.

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 fides parallel to one another; the remaining fides 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 are in a straight line.

Because AB is parallel to
DC, and the straight line
AC meets them, the al.
ternate angles BAC, ACD

D
are equal*; for the same
reason, the angle CDE is
equal to the angle ACD;
wherefore also BAC is e.
qual to CDE: And because

B

C E

the

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the triangles ABC, DCE have one angle at A equal to one at Book VI. D, and the sides about these angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangular to DCE: 66.6. Therefore the angle ABC is equal to the angle DCE : And the angle BAC was proved to be equal to ACD: Therefore the whole angle ACE is equal to the two angles ABC, BAC ; add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB : But ABC, BAC, ACB are equal to two right angles; therefore also the angles ACE, - 52. 1. ACB are equal to two right angles : And since at the point C, in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore a BC and d 14. I, CE are in a straight line. Wherefore, if two triangles, &c, Q. E. D.

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IN equal circles, angles, whether at the centers or cir- See N.

cumferences, have the same ratio which the circum. ferences on which they stand have to one another: So also have the sectors.

Let ABC, DEF be equal circles; and at their centers the angles BGC, EHF, and the angles BAC, EDF at their circumferences ; as the circumference BC to the circumference EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and also the sector BGC to the sector EHF.

Take any number of circumferences CK, KL, each equal to BC, and any number whatever FM, MN each equal to EF: And join GK, GL, HM, HN. Because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal * : Therefore what multiple foever the circum- a 27. 3o ference BL is of the circumference BC, the fame multiple is the angle BGL of the angle BGC: For the same reason, whatever multiple the circumference EN is of the circumference EF, the same multiple is the angle EHN of the angle EHF:

And

Book VI. And if the circumference BL be equal to the circumference

EN, the angle BGL is also equal to the angle EHN; and a 27. 3.

if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less : There being then four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF; of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and the angle BGL; and of the circumference EF, and of the angle EHF, any equimultiples what

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C 15. 6.

d 20. 3.

ever, viz. the circumference EN, and the angle EHN: And it has been proved, that, if the circumference BL be greater than EN, the angle BGL is greater than EHN; and if e.

qual, equal; and if less, less : As therefore the circumference bs. Def, s. BC to the circumference EF, so 6 is the angle BGC to the

angle EHF: But as the angle BGC is to the angle EHF, fo is the angle BAC to the angle EDF, for each is double of each: Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

Also, as the circumference BC to EF, fo is the sector BGC to the fector EHF. Join BC, CK, and in the circumferences BC, CK take any points X, O, and join BX, XC, CO, OK: Then, because in the triangles GBC, GCK the two fides BG,

GC are equal to the two CG, GK, and that they contain e¢ 4. I. qual angles; the base BC is equal to the base CK, and the

triangle GBC to the triangle GCK : And because the circumference BC is equal to the circumference CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the fame

circle: Wherefore the angle BXC is equal to the angle COK"; f 11. def. 3. and the segment BXC is therefore similar to the segment COK';

and

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