1Q V. And, ſuppoſe that as the Square C R is to the Square C P, ſo is the Line B N unto
another Line; which let be S X: And, as the Square C T is to the Square C Q ſo let F O
be to V Y. Now it is manifeſt, by the things which we have demonſtrated, in our Commentaries,
upon the fourth Propoſition of Archimedes, De Conoidibus & Spheæroidibus, that the
Square C P is equall to the Rectangle P S X; and alſo, that the Square C Q is equall to
the Rectangle Q V Y; that is, the Lines S X and V Y, are the Parameters of the Sections H S C
and M V C: But ſince the Triangles C P R and C Q T are alike; C R ſhall have to C P, the
ſame Proportion that C T hath to C Q: And, therefore, the (a) Square C R ſhall have
to the Square C P, the ſame proportion that the
50[Figure 50]
Square C T hath to the Square C Q: There
fore, alſo, the Line B N ſhall be to the Line
S X, as the Line F O is to V Y: But H C was
to C M, as A C to C E: And, therefore, alſo,
their halves C P and C Q, are alſo to one
another, as A D and E G: And. Permu
tando, C P is to A D, as C Q is to E G:
But it hath been proved, that A D is to B N,
as E G to F O; and B N to S X, as F O to
V Y: Therefore, exæquali, C P ſhall be
to S X, as C Q is to V Y. And, ſince the
Square C P is equall to the Rectangle P S X, and the Square C Q to the Rectangle Q V Y,
the three Lines S P, PC and S X ſhall be proportionalls, and V Q, Q C and V Y ſhal be
Proportionalls alſo: And therefore alſo S P ſhall be to P C as V Q to Q C And as P C
is to C H, ſo ſhall Q C. be to C M: Therefore, ex æquali, as S P the Diameter of the
Portion H S C is to its Baſe C H, ſo is V Q the Diameter of the portion M V S the
Baſe C M; and the Angles which the Diameter with the Baſes do contain, are equall; and the
Lines S P and V Q are parallel: Therefore the Portions, alſo, H S C and M V C ſhall be alike:
Which was propoſed to be demonſtrated
another Line; which let be S X: And, as the Square C T is to the Square C Q ſo let F O
be to V Y. Now it is manifeſt, by the things which we have demonſtrated, in our Commentaries,
upon the fourth Propoſition of Archimedes, De Conoidibus & Spheæroidibus, that the
Square C P is equall to the Rectangle P S X; and alſo, that the Square C Q is equall to
the Rectangle Q V Y; that is, the Lines S X and V Y, are the Parameters of the Sections H S C
and M V C: But ſince the Triangles C P R and C Q T are alike; C R ſhall have to C P, the
ſame Proportion that C T hath to C Q: And, therefore, the (a) Square C R ſhall have
to the Square C P, the ſame proportion that the
50[Figure 50]Square C T hath to the Square C Q: There
fore, alſo, the Line B N ſhall be to the Line
S X, as the Line F O is to V Y: But H C was
to C M, as A C to C E: And, therefore, alſo,
their halves C P and C Q, are alſo to one
another, as A D and E G: And. Permu
tando, C P is to A D, as C Q is to E G:
But it hath been proved, that A D is to B N,
as E G to F O; and B N to S X, as F O to
V Y: Therefore, exæquali, C P ſhall be
to S X, as C Q is to V Y. And, ſince the
Square C P is equall to the Rectangle P S X, and the Square C Q to the Rectangle Q V Y,
the three Lines S P, PC and S X ſhall be proportionalls, and V Q, Q C and V Y ſhal be
Proportionalls alſo: And therefore alſo S P ſhall be to P C as V Q to Q C And as P C
is to C H, ſo ſhall Q C. be to C M: Therefore, ex æquali, as S P the Diameter of the
Portion H S C is to its Baſe C H, ſo is V Q the Diameter of the portion M V S the
Baſe C M; and the Angles which the Diameter with the Baſes do contain, are equall; and the
Lines S P and V Q are parallel: Therefore the Portions, alſo, H S C and M V C ſhall be alike:
Which was propoſed to be demonſtrated
(a) By 22. of the
ſixth.
ſixth.
LEMMA. IV.
Let there be two Lines A B and C D; and let them be cut in the
Points E and F, ſo that as A E is to E B, C F may be to F D:
and let them be cut again in two other Points G and H; and
let C H be to H D, as A G is to G B. I ſay that C F ſhall be to
F H as A E is E G.
Points E and F, ſo that as A E is to E B, C F may be to F D:
and let them be cut again in two other Points G and H; and
let C H be to H D, as A G is to G B. I ſay that C F ſhall be to
F H as A E is E G.
For in regard that as A E is to E B, ſo is C F to F D; it followeth that, by Compounding,
as A B is to E B, ſo ſhall C D be to F D. Again, ſince that as A G is to G B, ſo is C H, to
H D; it followeth that, by Compounding and Converting, as G B is to A B, ſo ſhall H D be
51[Figure 51]
C D: Therefore, ex æquali, and Converting as E B
is to G B, ſo ſhall F D be to H D; And, by Conver
ſion of Propoſition, as E B is to E G, ſo ſhall F D
be to F H: But as A E is to E B, ſo is C F to F D:
Ex æquali, therefore, as A E is to E G, ſo
ſhall CF be to F H. Again, another way. Let
the Lines A B and C D be applyed to one another,
ſo as that they doe make an Angle at the parts A and C;
and let A and C be in one and the ſame Point: then
draw Lines from D to B, from H to G, and from F to E. And ſince that as A E is to E B,
ſo is C F, that is A F to F D; therefore F E ſhall be parallel to D B; (a) and likewiſe
H G ſhall be parallel to D B; for that A H is to H D, as A G to G B: (b) Therefore F E
and H G are parallel to each other: And conſequently, as A E is to E G, ſo is A H, that is,
C F to F H: Which was to be demonſtrated.
as A B is to E B, ſo ſhall C D be to F D. Again, ſince that as A G is to G B, ſo is C H, to
H D; it followeth that, by Compounding and Converting, as G B is to A B, ſo ſhall H D be
51[Figure 51]C D: Therefore, ex æquali, and Converting as E B
is to G B, ſo ſhall F D be to H D; And, by Conver
ſion of Propoſition, as E B is to E G, ſo ſhall F D
be to F H: But as A E is to E B, ſo is C F to F D:
Ex æquali, therefore, as A E is to E G, ſo
ſhall CF be to F H. Again, another way. Let
the Lines A B and C D be applyed to one another,
ſo as that they doe make an Angle at the parts A and C;
and let A and C be in one and the ſame Point: then
draw Lines from D to B, from H to G, and from F to E. And ſince that as A E is to E B,
ſo is C F, that is A F to F D; therefore F E ſhall be parallel to D B; (a) and likewiſe
H G ſhall be parallel to D B; for that A H is to H D, as A G to G B: (b) Therefore F E
and H G are parallel to each other: And conſequently, as A E is to E G, ſo is A H, that is,
C F to F H: Which was to be demonſtrated.