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composed of pyramids the bases of which are the aforesaid quadri- Book XII.
lateral 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. and the superficies of this folid polyhedron does -
not meet the lesser sphere in which is the circle FGH. for from the
point A draw k AZ perpendicular to the plane of the quadrilateral k. 11.11,
KBOS meetiog it in Z, and join BZ, ZK. and because AZ is per-
pendicular to the plane KBOS, it makes right angles with every
straight line meeting it in that plane; therefore AZ is perpendicular
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, 47. 1.
ZB are equal to the squares of AZ, ZK. take from these equals the
square of AZ, 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 manner 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 circie described from the center Z, and distance ZB
shall pass thro’ the points K,O,S, and KBOS shall 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 circumfe-
rences 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 circumference of the circle; there-
fore the circumference subtended by KB is greater than the fourth
part of the whole circumference of the circle KBOS, and consequent-
ly the angle BZK at the center is greater than a right angle. and
because the angle BZK is obtuse, the square of BK is greater i than 1. 12. 2.
the squares of BZ, ZK; that is, greater than twice the square of
BZ. Join KV, and because in the triangles KBV, OBV, KB, BV
are equal to OB, BV, and that they contain equal angles; the angle
KVB is equal to the angle OVB. and OVB is a right angle; m. 4. 1.
therefore also KVB is a right angle. and because BDis less than twice
DV, the rectangle contained by DB, BV is less than twice the rec-
tangle DVB; that is o, the square of KB is less than twice the 1.8. 6.
square of KV. but the square of KB is greater than twice the
square of BZ; therefore the square of KV is greater than the square
of BZ. and because BA is equal to AK, and that the squares of

Book XII. BZ, ZA are equal together to the square of BA, and the squares

of KV, VA to the square of AK; therefore the squares of BZ, ZA are equal to the squares of KV, VA; and of these the square of KV is greater than the square of BZ, therefore the square of VA is less than the square of ZA, and the straight line AZ greater than VA. much more then AZ is greater than AG, because in the preceding Proposition it was shewn that KV falls without the circle FGH. and AZ is perpendicular to the plane KBOS, and is therefore the Mortest of all the straight lines that can be drawn from A the center of the sphere to that planc. Therefore the plane KBOS does not meet the lesser sphere.

And that the other planes between the quadrants BX, KX fall without the lesser sphere, is thus demonstrated. from the point A draw AI perpendicular to the plane of the quadrilateral SOPT, and join 10; and as was demonstrated of the plane KBOS and the point 2, in the same way it may be shewn that the point I is the center of a circle described about SOPT, and that OS is greater than PT; and PT was shewn to be parallel to OS. therefore because the two trapeziums KBOS, SOPT inscribed in circles have their fides BK, OS parallel, as also OS, PT; and their other sides BO, KS, OP,

STall equal to one another, and that BK is greater than OS, and OS 0. 2. Lem. greater than PT, therefore the straight line ZB is greater than IO.

Join AO which will be equal to AB; and because AIO, AZB are
right angles, the squares of AI, 10 are equal to the square of AO
or of AB; that is, to the squares of AZ, ZB; and the square of
ZB is greater than the square of IO, therefore the square of AZ
is less than the square of Al; and the straight line AZ less than
the straight line Al. and it was proved that AZ is greater than AG;
much more then is AI greater than AG. therefore the plane SOPT
falls wholly without the lesser sphere. in the same manner it may
be demonstrated that the plane TPRY falls without the fame
sphere, as also the triangle YRX, viz. by the Cor. of 2d Lemma.
and after the same way it may be demonstrated that all the planes
which contain the solid polyhedron fall without the lesser sphere.
therefore in the greater of two spheres which have the same cen-
ter, a solid polyhedron is described the fuperficies of which does
not meet the lesser sphere. Which was to be done.

But the straight line AZ may be demonstrated to be greater than
AG otherwise and in a sorter manner, without the help of Prop.
86. as follows. From the point G draw GU at right angles to AG

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and join AU. if then the circumference BE be bisected, and its Book X!T.
half again bisected, and so on, there will at length be left a cir- u
cumference less than the circumference which is subtended by a
straight line cqual to GU inscribed in the circle BCDE. let this
be the circumference KB. therefore the straight line KB is less than
GU. and because the angle BZK is obtuse, as was proved in the
preceding, therefore BK is greater than BZ. but GU is greater
than BK; much more then is GU greater than BZ, and the square
of GU than the square of BZ. and AU is equal to AB; therefore
the square of AU, that is the squares of AG, GU are equal to the
square of AB, that is to the squares of AZ, ZB; but the square
of BZ is less than the square of GU; therefore the square of AZ
is greater than the square of AG, and the straight line AZ conse-
queatly greater than the straight line AG.

Cor. And if in the lesser sphere there be described a solid poly-
hedron by drawing straight lines betwixt the points in which the
Straight lines from the center of the sphere drawn to all the angles of
the folid polyhedron in the greater sphere meet the superficies of the
lesler ; in the fame order in which are joined the points in which the
fame lines from the center meet the superficies of the greater sphere;
the solid polyhedron in the sphere BCDE has to this other solid
lyhedron the triplicate ratio of that which the diameter of the sphere
BCDE has to the diameter of the other sphere. for if thefe two fo-
lid: be divided into the same number of pyramids, and in the same
order; the pyramids Mall be similar to one another, cach to each.
because they have the folid angles at their common vertex, the cen-
ter of the sphere, the same in each pyramid, and their other solid
angles at the bases equal to one another, each to each ?, because a. B. 11.
they are contained by three plane angles equal each to each; and the
pyramids are contained by the same number of similar planes; and
are therefore similar 6 to one another, each to each. but similar

py

b. 11. Def. ramids have to one another the triplicate ratio of their homologous sides, therefore the pyramid of which the base is the quadrilateral KBOS, and vertex A, has to the pyramid in the other sphere of the fame order, the triplicate ratio of their homologous fides; that is, of that ratio which AB from the center of the greater sphere has to the straight line from the same center to the superficies of the leffer sphere, and in like manner each pyramid in the greater sphere has to each of the same order in the lesser, the triplicate ratio of that which AB has to the femidiameter of the lesser sphere, and as one

po

c.Cor.8.12.

Book XII. antecedent is to its consequent, fo are all the antecedents to all the

consequents. Wherefore the whole solid polyhedron in the greater sphere has to the whole tolid polyhedron in the other, the triplicate ratio of that which AB the femidiameter of the first has to the femidiameter of the other ; that is, which the diameter BD of the greater has to the diameter of the other sphere.

PRO P. XVIII. THEOR,

SPHERES
PHERES have to one another the triplicate ratio

of that which their diameters have.

Let ABC, DEF be two spheres of which the diameters are BC, EF. the sphere ABC has to the sphere DEF the triplicate ratio of that which BC has to EF.

For if it has not, the sphere ABC shall have to a sphere either less or greater than DEF, the triplicate ratio of that which BC has to EF. First, let it have that ratio to a less, viz. to the sphere

GHK; and let the sphere DEF have the same center with GHK; 2. 17. 12. and in the greater sphere DEF describe : a folid polyhedron the

superficies of which does not meet the lefser sphere GHK; and in
А.

D
G

B

CEH

K

FM

N

the sphere ABC describe another similar to that in the sphere

DEF. therefore the solid polyhedron in the sphere ABC has to . Cor. 14. the solid polyhedron in the sphere DEF, the triplicate ratio b of

that which BC has to EF. but the sphere ABC has to the sphere GHK, the triplicate ratio of that which BC has to EF; therefore as the sphere ABC to the sphere GHK, fo is the folid polyhedron in the sphere ABC to the solid polyhedron in the sphere DEF. but the sphere ABC is greater than the folid polyhedron in it

therefore • also the sphere GHK is greater than the folid polyhe-Book XII. dron in the sphere DEF. but it is also less, because it is contained within it, which is imposible. therefore the sphere ABC has not to c. 14 S. any sphere less than DEF, the triplicate ratio of that which BC has to EF. In the same manner it may be demonstrated that the sphere DEF has not to any sphere less than ABC, the triplicate ratio of that which EF has to BC. Nor can the sphere ABC have to any sphere greater than DEF, the triplicate ratio of that which BC has to EF. for if it can, let it liave that ratio to a greater sphere LMN. therefore, by inversion, the fphere LMN has to the sphere ABC, the triplicate ratio of that which the diameter EF has to the diameter BC. but as the sphere LMN to ABC, so is the sphere DEF to some sphere, which must be less e than the sphere ABC, because c. 14. 5. the sphere LMN is greater than the sphere DEF. therefore the sphere DEF has to a sphere less than ABC the triplicate ratio of that which EF has to BC; which was shewn to be impossible. therefore the sphere ABC has not to any sphere greater than DEF the triplicate ratio of that which BC has to EF. and it was demonstrated that neither has it that ratio to any sphere less than DEF. Therefore the sphere ABC has to the sphere DEF, the sriplicate ratio of that which BC has to EF. 0. E. D.

FINI S.

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