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See N.

Which is the first proposition of the tenth book, and is necessary to

some of the propositions of this book. If from the greater of two unequal magnitudes, there be

taken more than its half, and from the remainder more than its half; and so on : there shall at length remain a magnitude less than the least of the proposed magnitudes.

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A

D

K

F

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Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more than its half, and from the remainder more than its half, and so on; there shall at length remain a magnitude less than C.

For C may be multiplied so as at length to become greater than AB. Let it be so multiplied, and let DE its multiple be greater than II AB, and let DE be divided into DF, FG, GE, each equal to C. From AB take BH greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: and let the divisions in AB be AK, KH, HB; and the divisions in DE be DF, FG, GE. And because DE is greater than AB, and that EG taken from DE is not greater than its half, but BH taken from AB is greater than its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than

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the half of GD, but HK is greater than the half of HA; therefore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C.

Q. E. D. And if only the halves be taken away, the same thing may in the same way be demonstrated.

PROP. I. THEOR.

F

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M
D

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Similar polygons inscribed in circles, are to one another

as the squares of their diameters. Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHKL; and let BM, GN be the diameters of the circles: as the square of BM is to the square of GN, so shall the polygon ABCDE be to the polygon FGHKL.

Join BE, AM, GL, FN: and because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and as BA to AE, so is GF to FL: therefore the two triangles BAE, GFL having one angle in one equal to one angle in the other, and the sides about the equal an

EG

L gles proportionals, are equiangular; and therefore the angle A EB is equal to the an

K gle FLG: but AEB is equal

* 21. 3. to AMB, because they stand upon the same circumference: and the angle FLG is, for the same son, equal to the angle FNG: therefore also the. angle AMB is equal to FNG: and the right angle BAM is equal to the right* angle GFN; wherefore • 31. 3. the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another : therefore as BM to GN, so * is BA to GF; and therefore • 4. 6. the duplicate ratio of BM to GN, is the same* with 10 Def. 5. the duplicate ratio of BA to GF: but the ratio of the & 22.5. square of BM to the square of GN, is the duplicate ratio of that which BM has to GN; and the ratio of the polygon ABCDE to the polygon FGHKL is the duplicate* of that which BA has to GF: therefore as * 20. 6. the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore, similar polygons, &c.

rea

*

* 20. 6.

Q. E. D.

PROP. II. THEOR.

See N.

Circles are to one another as the squares of their

diameters.

+ 6.4.

* 41.1.

Let ABCD, EFGH be two circles, and BD, FH their diameters: as the square of BD to the square of FH, so shall the circle ABCD be to the circle EFGH.

For, if it be not so, the square of BD must be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it I. First let it be to a space S less than the circle EFGH; and in the circle EFGH+ describe the square EFGH. This square is greater than half of the circle EFGH; because, if, through the points E, F, G, H, there be drawn tangents to the circle, the square EFGH is half* of the square described about the circle: and the circle is less than the square described about it; therefore the square EFGH is greater than half of the circle. Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE: therefore each of the triangles EKF, FLG, GMH, HNE, is greater than half of the segment of the circle in which it stands; because, if straight lines touching the circle be drawn through the points K, L, M, N, and the parallelograms upon the straight lines, EF, FG, GH, HE be completed, each of the triangles EKF, FLG, GMH, HNE, is the half* of the parallelogram in which it is : but every segment is less than the parallelogram in which it is; wherefore each of the triangles EKF, FLG, GMH, HNE, is greater than half the segment of the circle which contains it.

Again, if the remaining circumferences be divided each into two equal parts, and their extremities be joined by straight lines, by continuing to do this, there will at length

* 41.1.

$ 22.6.

For there is some square equal to the circle ABCD; let P be the side of it, and to three straight lines, BD, FH, and P, there can be a fourth proportional ; let this be Q: therefore + the squares of these four straight lines are proportionals; that is, to the squares of BD, FH, and the circle ABCD it is possible there may be a fourth proportional. Let this be S. And in like manner are to be understood some things in some of the following propositions.

1

1

remain segments of the circle, which together are less
than the excess of the circle EFGH above the space S;
because by the preceding Lemma, if from the greater
of two unequal magnitudes there be taken more
than its half, and from
the remainder more

X R than its half, and so

IN on, there shall at length B

F

H S remain a magnitude

L M less than the least of the proposed magnitudes. Let then the segments EK, KF, FL, LG, GM, MH, HN, NE, be those that remain, and are together less than the excess of the circle EFGH above S: therefore the rest of the circle, viz. the polygon EKFLGMHN is greater than the space S. Describe likewise in the circle ABCD the polygon AXBOCPDR similar to the polygon EKFLGMHN: as therefore the square of BD is to the square of FH *, so is the polygon AXBOCPDR to the polygon EKFLGMHN: but the square of BD is also to the square

of FHt, as the circle ABCD is to the space S; + Hyp. therefore as the circle ABCD is to the space S, so is * the * 11.5. polygon AXBOCPDR to the polygon EKFLGMHN: but the circle ABCD is greater than the polygon contained in it; wherefore the space S is greater than the polygon EKFLGMHN: but it is likewise less, as has been demonstrated; which is impossible. Therefore the square of BD is not to the square of FH, as the circle ABCD is to any space less than the circle EFGH. In the same manner, it may be demonstrated, that neither is the square of FH to the square of BD, as the circle EFGH is to any space less than the circle ABCD. Nor is the square of BD to the square of FH, as the circle ABCD is to any space greater than the circle EFGH. For if possible, let it be so to T, a space greater than the circle EFGH: therefore inversely, as the square of FH to

• 1. 12.

* 14. 5.

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the square of BD, so is the space T to the circle

&

* 14. 5.

as

ABCD: but as the space T is to the circle ABCD,
so is the circle EFGH to some space, which must be
less * than the circle ABCD, because the space T is
greater, by hypothesis, than the circle EFGH; there-
fore as the square of FH is to the square of BD, so is the
circle EFGH to a space less than the circle ABCD,
which has been demonstrated to be impossible: there-
fore the
square
of BD is not to the

square
of FH

the circle ABCD is to any space greater than the circle EFGH: and it has been demonstrated that neither is the

square of BD to the square of FH, as the circle ABCD to any space less than the circle EFGH: wherefore, as the square of BD to the square of FH, so is the circle ABCD to the circle EFGH. Circles therefore are, &c. Q. E. D.

PROP. III. THEOR.

See N.

Every pyramid having a triangular base, may be divided

into two equal and similar pyramids having triangular bases, and which are similar to the whole pyramid; and into two equal prisms which together are greater than half of the whole pyramid.

Let there be a pyramid of which the base is the
triangle ABC and its vertex the point D: the pyramid
ABCD may be divided into two equal and
similar pyramids having triangular bases, D
and similar to the whole; and into two equal
prisms which together shall be greater than
half of the whole pyramid.

K L
Divide AB, BC, CA, AD, DB, DC, each
into two equal parts in the points E, F,
G, H, K, L, and join EH, EG, GH,
HK, KL, LH, EK, KF, FG. Because
AE is equal to EB, and AH to HD, HE

B

# For as, in the foregoing note at *, it was explained how it was possible, there could be a fourth proportional to the squares of BD, FH, and the circle ABCD, which was named S; so, in like manner, there can be a fourth proportional to this other space, named T, and the circles ABCD, EFGH. And the like is to be understood in some of the following propositions.

$ Because, as a fourth proportional to the squares of BD, FH, and the circle ABCD, is possible, and that it can neither be less nor greater than the circle EFGH, it must be equal to it.

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