37(25)
which RUM by Lemma II.
is equal to a rectangle under UO and a line
drawn through the points U and O to the further ſurface of the ſphere YN.
Therefore the point Q is in the ſurface of the ſphere YN; it is therefore
common to the ſpheres YN and OTS; and I ſay that theſe ſpheres touch in
the ſaid point Q. For from the point U let a line UZ be drawn in any
plane of the ſphere OTS, and being produced let it cut the three
ſpheres in the points Z, D, H, K, P, B. The rectangle ZUB in the
ſphere OTS is by Lemma I. and II. equal to the rectangle DUP terminated
by the ſpheres XM and YN. But DU is greater than ZU, becauſe the
ſpheres XM and OTS touch in the point O, and therefore any other line from
U but UO muſt meet the ſurface of OTS before it meets the ſurface XM.
Since then ZUB = DUP, and DU is greater than ZU, UP muſt be leſs
than UB, and the point B will fall without the ſphere YN; and by the
fame reaſon, all other points in the ſurface of the ſphere OTS, except the
point Q.
drawn through the points U and O to the further ſurface of the ſphere YN.
Therefore the point Q is in the ſurface of the ſphere YN; it is therefore
common to the ſpheres YN and OTS; and I ſay that theſe ſpheres touch in
the ſaid point Q. For from the point U let a line UZ be drawn in any
plane of the ſphere OTS, and being produced let it cut the three
ſpheres in the points Z, D, H, K, P, B. The rectangle ZUB in the
ſphere OTS is by Lemma I. and II. equal to the rectangle DUP terminated
by the ſpheres XM and YN. But DU is greater than ZU, becauſe the
ſpheres XM and OTS touch in the point O, and therefore any other line from
U but UO muſt meet the ſurface of OTS before it meets the ſurface XM.
Since then ZUB = DUP, and DU is greater than ZU, UP muſt be leſs
than UB, and the point B will fall without the ſphere YN; and by the
fame reaſon, all other points in the ſurface of the ſphere OTS, except the
point Q.
The Demonſtration is ſimilar and equally eaſy in all caſes, whether the
ſpheres touch exlernally or internally.
ſpheres touch exlernally or internally.
LEMMA IV.
Let there be a plane AC, and a ſphere FGD through whoſe center O let
FODB be drawn perpendicular to the plane, and from F any right line
FGA cutting the ſphere in G and the plane in A; I ſay that the rectangle
AFG = the rectangle BFD. For let the given ſphere and plane be cut by
the plane of the triangle ABF, the ſection of the one will be the circle GDF,
and of the other the right line ABC. Since the line FB is perpendicular
to the plane AC, it will be alſo to the right line AC. Having then a circle
FDB, and a right line AC in the ſame plane; and a line FDB paſſing thro'
the center perpendicular to AC, join D and G, and in the quadrilateral
figure ABDG the angles at B and G being both right ones, it will be in a
circle, and the rectangle AFG = the rectangle BFD; and the ſame may be
proved in any other ſection of the ſphere.
FODB be drawn perpendicular to the plane, and from F any right line
FGA cutting the ſphere in G and the plane in A; I ſay that the rectangle
AFG = the rectangle BFD. For let the given ſphere and plane be cut by
the plane of the triangle ABF, the ſection of the one will be the circle GDF,
and of the other the right line ABC. Since the line FB is perpendicular
to the plane AC, it will be alſo to the right line AC. Having then a circle
FDB, and a right line AC in the ſame plane; and a line FDB paſſing thro'
the center perpendicular to AC, join D and G, and in the quadrilateral
figure ABDG the angles at B and G being both right ones, it will be in a
circle, and the rectangle AFG = the rectangle BFD; and the ſame may be
proved in any other ſection of the ſphere.
LEMMA V.
Let there be a plane ABD and a ſphere EGF, through whoſe center O
let FOEC be drawn perpendicular to the plane, and in any other plane let
FHI be drawn ſo that the rectangle IFH = the rectangle CFE: if
let FOEC be drawn perpendicular to the plane, and in any other plane let
FHI be drawn ſo that the rectangle IFH = the rectangle CFE: if