112[35]
Epitagma III.
There are here but four Caſes, viz.
when the order of the
given points is A, E, I, U; A, I, E, U; E, A, U, I; or E, U, A, I;
the two ſirſt of theſe are not poſſible unleſs the given ratio be the ratio of a
greater to a leſs; nor the two latter, unleſs it be of a leſs to a greater, and
as theſe are reduced to the ſirſt two by reading every where E for A, I
for U, and the contrary, I ſhall omit ſpecifying them.
given points is A, E, I, U; A, I, E, U; E, A, U, I; or E, U, A, I;
the two ſirſt of theſe are not poſſible unleſs the given ratio be the ratio of a
greater to a leſs; nor the two latter, unleſs it be of a leſs to a greater, and
as theſe are reduced to the ſirſt two by reading every where E for A, I
for U, and the contrary, I ſhall omit ſpecifying them.
Case I.
If the order of the given points be A, E, I, U, the conſtruc-
tion will be effected by Fig. 44, wherein B is made to fall beyond I, and
C beyond E, and DH is drawn parallel to BC. That O, when this con-
ſtruction is uſed, will fall between E and I, is eaſily made appear by rea-
ſoning in a manner ſimilar to what was done in Caſe V. of Epitagma I.
tion will be effected by Fig. 44, wherein B is made to fall beyond I, and
C beyond E, and DH is drawn parallel to BC. That O, when this con-
ſtruction is uſed, will fall between E and I, is eaſily made appear by rea-
ſoning in a manner ſimilar to what was done in Caſe V. of Epitagma I.
Case II.
The conſtruction of this Caſe, where the order of the points
is A, I, E, U, is not materially different from that above exhibited as ap-
pears by Fig. 45, and that O will fall between I and E is manifeſt without
farther illuſtration.
is A, I, E, U, is not materially different from that above exhibited as ap-
pears by Fig. 45, and that O will fall between I and E is manifeſt without
farther illuſtration.
Limitation.
In theſe two Caſes the given ratio of R to S cannot be
leſs than that which the ſquare on AU bears to the ſquare on a line which
is the difference of two mean proportionals between AI and EU, AE and
IU. For by Lemma I. the leaſt ratio which the rectangle contained by AO
and UO can have to the rectangle contained by EO and IO; or, which
is the ſame thing, that R can have to S, will be when the point O is
the interſection of the diameter AU, of a circle AYUV, with a ſtraight
line YV. joining the tops of two perpendiculars EV, IY to the indeſi-
nite line, on contrary ſides thereof, and terminating in the periphery of
the circle. Produce VE (Fig. 27.) to meet the circle again in K, and
draw the diameter KL; join LY and KY, on which, produced, let fall
the perpendicular UF. Now, ſince by Lemma III. KF is a mean propor-
tional between AI and EU, and YF a mean proportional between AE
and IU: it remains only to prove that the ratio of the rectangle con-
tained by AO and OU to the rectangle contained by EO and OI is the
ſame with the ratio which the ſquare on AU bears to the ſquare on KY,
which is the diſſerence between KF and YF. Becauſe the angles E and
KYL are both right, and the angles EVO and KYL equal (Eu. III. 21.)
the triangles EVO and YLK are ſimilar; and ſo VO is to EO as AU (LK)
is to KY; or the ſquare on VO is to the ſquare on EO as the ſquare on AU
is to the ſquare on KY. Now the triangles EVO, IYO being alſo
leſs than that which the ſquare on AU bears to the ſquare on a line which
is the difference of two mean proportionals between AI and EU, AE and
IU. For by Lemma I. the leaſt ratio which the rectangle contained by AO
and UO can have to the rectangle contained by EO and IO; or, which
is the ſame thing, that R can have to S, will be when the point O is
the interſection of the diameter AU, of a circle AYUV, with a ſtraight
line YV. joining the tops of two perpendiculars EV, IY to the indeſi-
nite line, on contrary ſides thereof, and terminating in the periphery of
the circle. Produce VE (Fig. 27.) to meet the circle again in K, and
draw the diameter KL; join LY and KY, on which, produced, let fall
the perpendicular UF. Now, ſince by Lemma III. KF is a mean propor-
tional between AI and EU, and YF a mean proportional between AE
and IU: it remains only to prove that the ratio of the rectangle con-
tained by AO and OU to the rectangle contained by EO and OI is the
ſame with the ratio which the ſquare on AU bears to the ſquare on KY,
which is the diſſerence between KF and YF. Becauſe the angles E and
KYL are both right, and the angles EVO and KYL equal (Eu. III. 21.)
the triangles EVO and YLK are ſimilar; and ſo VO is to EO as AU (LK)
is to KY; or the ſquare on VO is to the ſquare on EO as the ſquare on AU
is to the ſquare on KY. Now the triangles EVO, IYO being alſo