a.16.3.

Book. XII. circle draw GA at right angles to BD, and produce it to C; therefore AC touches a the circle EFGH: Then, if the circumference BAD be bifected, and the half of it be again bisected, and so on, there muft at length remain a circumference lefs b than AD: Let this be LD: and

b Lemma.

a 21 3.

:

D

from the point L draw LM per-
pendicular to BD, and produce
it to N and join LD, DN.
Therefore LD is equal to DN;
and because LN is parallel to AC B
and that AC touches the circle
EFGH; therefore LN does not
meet the circle EFHG: And much
lefs fhall the ftraight lines LD,
DN meet the circle EFGH: So
that if ftraight lines equal to LD be applied in the circle
ABCD from the point L around to N, there fhall be defcribed
in the circle a polygon of an even number of equal fides not
meeting the leffer circle. Which was to be done,

LEMMA II.

I
TF two trapeziums ABCD, EFGH be infcribed in
the circles, the centres of which are the points K,
L; and if the fides AB, DC be parallel, as alfo EF,
HG; and the other four fides AD, BC, EH, FG, be
all equal to one another; but the fide AB greater
than EF, and DC greater than HG. The ftraight
line KA from the centre of the circle in which the
greater fides are, is greater than the ftraight line LE
drawn from the centre to the circumference of the
other circle.

:

If it be poffible, let KA be not greater than LE; then KA must be either equal to it, or lefs. First, let KA be equal to LE Therefore becaufe in two equal circles, AD, BC in the one are equal to EG, FG in the other, the circumferences AD, BC are equal a to the circumferences EH, FG; but becaufe the ftraight lines AB, DC are refpectively 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 alfo equal to it, which is impoffible:

impoffible Therefore the ftraight line KA is not equal to Book XII. LÉ.

But let KA be less than LE, and make LM equal to KA, and from the centre L, and diftance LM defcribe the circle MNOP, meeting the ftraight line LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM which are respectively parallel a to, and lefs than EF, FG, GH, HE: Then, be-a 2. 6. caufe EH is greater than MP, AD is greater than MP; and

H

K

M N

BE

the circles ABCD, MNOP are equal; therefore the circumference AD is greater than MP; for the fame reason, the circumference BG is greater than NO; and because 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 fame reason, the circumference DC is greater than PO: Therefore the whole circumference ABCD is greater than the whole MNOP; but it is likewife equal to it, which is impoflible: Therefore KA is not lefs than LE; nor is it equal to it; the ftraight line KA muft therefore be greater than LE. Q. E. D.

COR. And if there be an ifofceles triangle the fides of which are equal to AD, BC, but its base less than AB the greater of the two fides AB, DC: the ftraight line KA may, in the fame manner, be demonftrated to be greater than the ftraight line drawn from the centre to the circumference of the circle defcribed about the triangle.

PROP.

Book XII.

See N.

a 15. 3.

b 16. 12.

€ 18. 11.

Th

PROP. XVII. PRO B.

10 defcribe in the greater of two spheres which have the fame centre, a folid polyhedron, the fuperfices of which fhall not meet the leffer sphere.

Let there be two fpheres about the fame centre A; it is required to defcribe in the greater a folid polyhedron, the superficies of which fhall not meet the leffer fphere.

Let the fpheres be cut by a plane paffing through the centre; the common fections of it with the fpheres fhall be circles; because the sphere is defcribed by the revolution of a femicircle about the diameter remaining unmoveable; fo that in whatever pofition the femicircle be conceived, the common fection of the plane in which it is with the fuperficies of the sphere is the circumference of a circle; and this is a great circle of the fphere, because the diameter of the sphere, which is likewise the diameter of the circle, is greater a than any straight line in the circle or sphere: Let then the circle made by the fection of the place with the greater fphere BCDE, and with the leffer fphere FGH; and draw the two diameters BD, CE, at right angles to one another; and in BCDE, the greater of the two circles, defcribeba polygon of an even number of equal fides not meeting the lefler circle FGH; and let its fides, in BE the fourth part of the circle, be BK, KL, LM, ME; join KA and produce it to N; and from A draw AX at right angles to the plane of the circle BCDE meeting the fuperficies of the sphere in the point X; and let planes pafs through AX and each of the ftraight lines BD, KN, which, from what has been faid, shall produce great circles on the fuperficies of the fphere, and let BXD, KXN be the femicircles thus made upon the diameters BD, KN: Therefore because XA is at right angles to the plane of the circle BCDE, every plane which paffes through XA is at right angles to the plane of the circle BCDA; wherefore the femicircles BXD, KXN are at right angles to the plane: And because the femicircles BED, BXD, KXN, upon the equal diameters BD, KN, are equal to one another, their halves BE, BX, KX, are equal to one another: Therefore, as many fides of the polygon as are in BE, fo many there are in BX, KX equal to the fides BK, KL, LM, ME: Let thefe polygons be defcribed, and their fides BO, OP, PR, RX; KS, ST, TY, YX, and join

OS,

D

OS, PT. RY, and from the points, O, S draw OV, SQ perpen- Book XII. diculars to AB, AK: And because the plane BOXD is at right angles to the plane BCDE, and in one of them BOXD, ÖV is drawn perpendicular to AB the common fection of the planes, therefore OV is perpendicular a to the plane BCDE: For the a 4. def. 11. fame reafon SQ is perpendicular to the fame plane, becaufe the plane KSXN is at right angles to the plane BCDE. Join VQ; and because in the equal femicircles BXD, KXN the

B

circumferences BO, KS are equal, and OV, SQ are perpendicular to their diameters, therefore d OV is equal to SQ, d 26. 1. and BV equal to KQ. But the whole BA is equal to the whole KA, therefore the remainder VA is equal to the remainder QA. As therefore BV, is to VA, fo is KQ to QA, wherefore VQ is parallel e to BK: And because OV, SQ are each of e z. 6. them at right angles to the plane of the circle BCDE, OV is parallel f to SQ; and it has been proved that it is alfo equal f6.. 11. to it; therefore QV, SO are equal and parallel 8: And becaufe g. 33 1., QV is parallel to SO, and alfo to KB: OS is parallel h to BK; h 9. 11. and therefore BO, KS which join them are in the fame plane

to

Book XII. in which thefe 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 ftraight lines AB, AK it may be demonftrated that TP is parallel to KB in the very fame way that SO was fhown to be parallel to the fame KB; wherefore a TP is parallel to SO, and the quadrilateral figure SOPT is in one plane: For the fame reafon, the quadrilateral TPRY is in one plane: And the figure YRX is alfo in one plane b

2.9. 11.

b 2.11.

Therefore, if from the points O, S, P, T, R, Y there be drawn ftraight lines to the point A, there, fhall be formed a folid polyhedron between the circumferences BX, KX compofed of pyramids the bafes 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 fame conftruction be made upon each of the fides KL, LM, ME, as has been done upon BK, and the like be done alfo in the other three quadrants, and in the other hemifphere; there fhall be formed a folid polyhedron defcribed in the fphere, compof

ed