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In the greater of two circles that have the same centre, to

inscribe a polygon of an even number of equal sides, that shall not meet the lesser circle.

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Let ABCD, EFGH be two given circles having the same centre K. it is required to inscribe in the greater circle ABCD, a polygon of an even number of equal sides, that shall not meet the lesser circle.

Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the lesser circle draw GA at right angles to BD, and produce it to C; therefore AC touches * the circle Cor. 16, EFGH: then, if the circumference BAD be bisected, 3. and the half of it be again bisected, and so on, there must at length remain a circumference less * than AD: * Lemma. let this be LD; and from the point L draw LM perpendicular to BD, and produce it to N; and join LD, DN: therefore LD is equal to DN: and because LN is parallel to AC, and that AC touches the LEL_XGMD circle EFGH; therefore LN does not meet the circle EFGH; and much less shall the straight lines LD, DN, meet the circle EFGH: so that if straight lines equal to LD be applied in the circle ABCD from the point L around to N, there shall be inscribed in the circle a polygon of an even number of equal sides not meeting the lesser circle. Which was to be done.

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LEMMA II.

If two trapeziums ABCD, EFGH be inscribed in the

circles, the centres of which are the points K, L; and if the sides, AB, DC be parallel, as also EF, HG; and the other four sides AD, BC, EH, FG, be all equal to one another ; but the side AB greater than EF, and DC greater than HG: the straight line KA from the centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre to the circumference of the other circle.

If it be possible, let KA be not greater than LE; then KA must be either equal to it, or less than it. First, let KA be equal to LE: therefore, because in

28. 3.

* 2. 6.

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two equal circles AD, BC, in the one, are equal to EH, FG in the other, the circumferences AD, BC, are equal * to the circumferences EH, FG; but because the straight lines AB, DC are respectively greater than EF, GH, the circumferences AB, DC are greater than EF, HG; therefore the whole circumference ABCD is greater than the whole EFGH: but it is also equal to it, which is impossible: therefore the straight line KA is not equal to LE.

But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM which are respectively parallel * to and less than EF, FG, GH, HE: then because EH is greater than MP, AD is greater than MP;, and the circles ABCD, MNOP are equal; therefore the circunference AD is greater than MP: for the same reason the cireumference BC is greater

HAG than NO: and be

PO cause the straight line AB is greater than EF, which is greater than MN, much more is AB greater than MN: therefore the circumference AB is greater than MN; and for the same reason, the cirference DC is greater than PO: therefore the whole circumference ABCD is greater than the whole MNOP: but it is likewise equal to it, which is impossible; therefore KA is not less than LE: nor is it equal to it; therefore the straight line KA must be greater than LE.

Cor. And if there be an isosceles triangle, the sides of which are equal to AD, BC, but its base less than AB the greater of the two sides AB, DC; the straight line KA

may, in the same manner, be demonstrated to be greater than the straight line drawn from the centre to the circumference of the circle described about the triangle.

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In the greater of two spheres which have the same centre, See N.

to inscribe a solid polyhedron, the superficies of which
shall not meet the lesser sphere.

Let there be two spheres about the same centre A: it is required to inscribe in the greater a solid polyhedron, the superficies of which shall not meet the lesser sphere.

Let the spheres be cut by a plane passing through the centre; the common sections of it with the spheres shall be circles; because the sphere is described by the revolution of a semicircle about the diameter remaining unmoveable; so that in whatever position the semicircle be conceived, the common section of the plane in which it is with the superficies of the sphere is the circumference of a circle; and this is a great circle of the sphere, because the diameter of the sphere, which is likewise the diameter of the circle, is greater * than * 15. 3. any straight line in the circle or sphere. Let then the circle made by the section of the plane with the greater sphere be BCDE, and with the lesser sphere be FGH; and draw the two diameters BD, CE, at right angles to one another; and in BCDE, the greater of the two circles, inscribe * a polygon of an even number of * 16. 12. equal sides not meeting the lesser circle FGH; and let its sides, in BE the fourth part of the circle, be BK, KL, LM, ME; join KA, and produce it to N; and from A drawt AX at right angles to the plane of the + 12. 11. circle BCDE, meeting the superficies of the sphere in the point X: and let planes pass through AX, and each of the straight lines BD, KN, which, from what has been said, shall produce great circles on the superficies of the sphere, and let BXD, KXN be the semicircles thus made upon the diameters BD, KN: therefore, because XA is at right angles to the plane of the circle BCDE, every plane which passes through XA is at right * angles to the plane of the circle BCDE; where- * 18. 11. fore the semicircles BXD, KXN are at right angles to that plane: and because the semicircles BED, BXD, KXÑ upon the equal diameters BD, KN, are equal to one another, their halves BE, BX, KX, are equal to one another; therefore as many sides of the polygon

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B

H

* 26.1.

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as are in BE, so many are there in BX, KX, equal to the sides BK, KL, LM, ME: let these polygons be described, and their sides be BO, OP, PR, RX; KS, ST, TY, YX; and join OS, PT, RY; and from the points 0, S, draw OV, SQ perpendiculars to AB, AK: and because the plane BOXD is at right angles to the plane BCDE, and in one of them BOXD, OV is

drawn perpendicular to AB the common section * 4 Def. 11. of the planes, therefore OV is perpendicular * to the

plane BCDE: for the same reason SQ is perpendicu-
lar to the same plane, because the plane KSX is at
right angles to the plane BCDE. Join VQ: and because
in the equal semi-
circles BXD, KXN,
the circumferences
BO, KS are equal,
and OV, SQ are per-
pendicular to their
diameters, therefore*
OV is equal to SQ,
and BV equal to
KQ: but the whole
BA is equal to the
whole KA, therefore
the remainder VA
is equal to the re-
mainder QA: therefore as BV is to VA, so is KQ to
QA; wherefore VQ is parallel * to BK: and because
OV, SQ are each of them at right angles to the plane
of the circle BCDE, OV is parallel * to SQ; and
it has been proved, that it is also equal to it; therefore
QV, SO are equal * and parallel: and because QV is
parallel to SO, and also to KB; OS is parallel * to BK;
and therefore BO, KS, which join them are in the same
plane in which these parallels are, and the

quadrilateral figure KBOS is in one plane: and if PB, TK be joined, and perpendiculars be drawn from the points P, T to the straight lines AB, AK, it may be demonstrated, that TP is parallel to KB in the very same way that SO was shewn to be parallel to the same KB; wherefore * TP is parallel to SO, and the quadrilateral figure SOPT is in one plane: for the same reason the quadrilateral TPRY is in one plane: and the figure YRX * is also in one plane: therefore, if from the points O, S, P, T, R, Y, there be drawn straight lines to the

** 2.6.

* 6. 11.

* 33. 1.

9. 11.

* 9. 11.

1

* 2. 11.

point A, there will be formed a solid polyhedron be-
tween the circumferences BX, KX, composed of pyra-
mids, the bases of which are the quadrilaterals KBOS,
SOPT, TPRY, and the triangle YRX, and of which
the common vertex is the point A: and if the same
construction be made upon each of the sides KL, LM,
ME, as has been done upon BK, and the like be done
also in the other three quadrants, and in the other
hemisphere; there will be formed a solid polyhedron
inscribed in the sphere, composed of pyramids, the bases
of which are the aforesaid quadrilateral figures, and
the triangle YRX, and those formed in the like manner
in the rest of the sphere, the common vertex of them all
being the point A.

Also the superficies of this solid polyhedron, shall
not meet the lesser sphere in which is the circle FGH.
For, from the point A draw * AZ perpendicular to the * 11.11.
plane of the quadrilateral KBOS, meeting it in Z, and
join BZ, ZK: and because AZ is perpendicular to the
plane KBOS, it makes right angles with every straight
line meeting it in that plane; therefore AZ is perpen-
dicular to BZ and ZK: and because AB is equal to
AK, and that the squares of AZ, ZB are equal to the
square of AB, and the squares of AZ, ZK, to the
square of AK *; therefore the squares of AZ, ZB, are • 47, 1.
equal to the squares of AZ, ZK: take from these equals
the square of AZ, and the remaining square of BZ is
equal to the remaining square of ZK; and therefore
the straight line BZ is equal to ZK: in the like man-
ner it may be demonstrated that the straight lines
drawn from the point Z to the points O, S, are equal
to BZ or ZK; therefore the circle described from the
centre Z, and distance ZB will pass through the points
K, O, S, and KBOS will be a quadrilateral figure in
the circle: and because KB is greater than QV, and
QV equal to SO, therefore KB is greater than SO:
but KB is equal to each of the straight lines BO, KS;
wherefore each of the circumferences cut off by KB,
BO, KS, is greater than that cut off by OS; and these
three circumferences, together with a fourth equal to
one of them, are greater than the same three together
with that cut off by OS; that is, than the whole cir-
cumference of the circle; therefore the circumference
subtended by KB is greater than the fourth part of
the whole circumference of the circle KBOS, and con-

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