The Elements of Euclid; viz. the first six books, together with the eleventh and twelfth. Also the book of Euclid's Data. By R. Simson. To which is added, A treatise on the construction of the trigonometrical canon [by J. Christison] and A concise account of logarithms [by A. Robertson].

FGHKL is the duplicate of that which BA has to GF: 20. 6. therefore as the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore, similar polygons, &c.

Q. E. D.

PROPOSITION II.

THEOR.-Circles are to one another as the squares of their See N. diameters.

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 n.ust be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it . First let it be to a space S less than the circle EFGH; and in the circle EFGH † † 6. 4. describe the square EFGHI. 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: is less than the square described about it square EFGH is greater than half of the circle circumferences EF, FG, GH, HE, each into two

and the circle 41. 1. therefore the

Divide the

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,

R K

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 seg-41. I. ment is less than the parallelogram in which it is: wherefore

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 t† the squares of these four straight lines are † 22, 6. 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.

• J. 12.

+ Hyp.

11. 5.

⚫ 14. 5.

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 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 than its half, and so on, there shall at length remain a
magnitude 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. De-
scribe 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 FH †, as the circle ABCD is to the space S;
therefore as the circle ABCD is to the space S, so is * the poly-
gon AXBOCPDR to the polygon EKFLGMHN: but the cir-
cle ABCD is greater than the polygon contained in it; where-
fore the s is
space 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

the circle ABCD

For if possi

is to any space greater than the circle EFGH.
ble, let it be so to T, a space greater than the circle EFGH:
therefore inversely, as the square of FH to the square of BD, so
is the space T to the circle ABCD: but as the space T is

For as, in the foregoing note, it was explained how it was possible there could be a fourth proportional to the squares of BD, FH, and the circle

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 14. 5. greater, by hypothesis, than the circle EFGH: therefore 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: therefore the square of BD is not to the square of FH, as 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,

PROPOSITION III.

THEOR.-Every pyramid having a triangular base, may be See N. 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, and similar to the whole; and into two equal prisms which together shall be greater than half of the whole pyramid.

1 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 is parallel to DF: for the same reason, HK is parallel to AB; therefore HEBK is a parallelogram, and HK equal to EB: but EB is equal to AE; therefore also AE is 34. 1. equal to HK: and AH is equal to HD; wherefore EA, AH are equal to KH, HD, each to each; and the angle EAH is equal*

ABCD, which was named S; so, in like manner, there can be a fourth propor tional 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 eircle EFGH, it must be equal to it.

* 2.6.

29. 1.

* 4. 1.

10. 11.

4. 1.

• C. 11.

4.6.

* 21. 6.

* B. 11. & 11 Def. 11.

41. 1.

.40.11.

to the angle KHD; therefore the base EH is equal to the base
KD, and the triangle AEH equal and similar to the triangle
HKD: for the same reason, the triangle AGH is equal and
similar to the triangle HLD. Again, because the two straight
lines EH, HG, which meet one another, are parallel to KD, DL,
that meet one another, and are not in the same plane with
them, they contain equal angles; therefore the angle EHG
is equal to the angle KDL: and because EH,
HG are equal to KD, DL, each to each, and the
angle EHG equal to the angle KDL; therefore
the base EG is equal to the base KL, and the
triangle EHG equal and similar to the trian-
gle KDL: for the same reason, the triangle AEG
is also equal and similar to the triangle HKL:
therefore the pyramid, of which the base is the
triangle AEG, and of which the vertex is the
point H, is equal* and similar to the pyramid, the base of
which is the triangle KHL, and vertex the point D. And
because HK is parallel to AB, a side of the triangle ADB, the
triangle ADB is equiangular to the triangle HDK, and their
sides are proportionals; therefore the triangle ADB is similar
to the triangle HDK: and for the same reason, the triangle
DBC is similar to the triangle DKL; and the triangle ADC to
the triangle HDL; and also the triangle ABC to the triangle
AEG: but the triangle AEG is similar to the triangle HKL,
as before was proved; therefore the triangle ABC is similar
to the triangle HKL: and therefore the pyramid of which the
base is the triangle ABC, and vertex the point D, is similar *
to the pyramid of which the base is the triangle HKL, and
vertex the same point D: bnt the pyramid of which the base
is the triangle HKL, and vertex the point D, is similar, as
has been proved, to the pyramid the base of which is the
triangle AEG, and vertex the point H; wherefore the pyra-
mid, the base of which is the triangle ABC, and vertex the
point D, is similar to the pyramid of which the base is the
triangle AEG, and vertex H: therefore each of the pyramids
AEGH, HKLD is similar to the whole pyramid ABCD. And
because BF is equal to FC, the parallelogram EBFG is double*
of the triangle GFC: but when there are two prisms of the
same altitude, of which one has a parallelogram for its base,
and the other a triangle that is half of the parallelogram, these
prisms are equal to one another; therefore the prism having

the parallelogram EBFG for its base, and the straight line KH opposite to it, is equal to the prism having the triangle GFC for its base, and the triangle HKL opposite to it; for they are of the same altitude, because they are between the parallel * planes ABC, HKL: and it is manifest that each of 15. 11. these prisms is greater than either of the pyramids of which the triangles AEG, HKL are the bases, and the vertices the points, H, D; because, if EF be joined, the prism having the parallelogram EBFG for its base, and KH the straight line opposite to it, is greater than the pyramid of which the base is the triangle EBF, and vertex the point K: but this pyramid is equal to the pyramid, the base of which is the C. 11.. triangle AEG, and vertex the point H; because they are contained by equal and similar planes: wherefore the prism having the parallelogram EBFG for its base, and opposite side KH, is greater than the pyramid of which the base is the triangle AEG, and vertex the point H: and the prism of which the base is the parallelogram EBFG, and opposite side KH, is equal to the prism having the triangle GFC for its base, and HKL the triangle opposite to it; and the pyramid of which the base is the triangle AEG, and vertex H, is equal to the pyramid of which the base is the triangle HKL, and the vertex D: therefore the two prisms before mentioned are greater than the two pyramids of which the bases are the triangles AEG, HKL, and vertices the points H, D. Therefore, the whole pyramid of which the base is the triangle ABC, and vertex the point D, is divided into two equal pyramids similar to one another, and to the whole pyramid; and into two equal prisms; and the two prisms are together greater than half of the whole pyramid. Q. E. D.

PROPOSITION IV.

THEOR.-If there be two pyramids of the same altitude upon See N. triangular bases, and each of them be divided into two equal pyramids similar to the whole pyramid, and also into two equal prisms; and if each of these pyramids be divided in the same manner as the first two, and so on; as the base of one of the first two pyramids is to the base of the other, so shall all the prisms in one of them be to all the prisms in the other, that are produced by the same number of divisions.

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The Elements of Euclid; viz. the first six books, together with the eleventh ...