104[27]
quently, by compound ratio, the rectangle contained by LS and SM is to
that contained by SE and SI as the rectangle contained by VO and OY,
or its equal, the rectangle contained by AO and OU is to that contained
by EO and IO. Now the triangles VSL and NSR having the angles at
R and L equal, and the angle at S common, are ſimilar; and therefore SR
is to SN as SL is to SV; conſequently, the rectangle contained by SR and
SV, or its equal, the rectangle contained by SA and SU is equal to that
contained by SN and SL: but SN is neceſſarily greater than SM, in con-
ſequence whereof the rectangle contained by SN and SL, or its equal, the
rectangle contained by AS and SU is greater than that contained by SM
and SL; wherefore the ratio which the rectangle AS, SU bears to the rect-
angle ES, SI is greater than that which the rectangle SM, SL bears to it,
and of courſe, greater than the ratio which the rectangle AO, UO bears to
the rectangle EO, IO; and that, on which ſide ſoever of the point O, S is
taken.
that contained by SE and SI as the rectangle contained by VO and OY,
or its equal, the rectangle contained by AO and OU is to that contained
by EO and IO. Now the triangles VSL and NSR having the angles at
R and L equal, and the angle at S common, are ſimilar; and therefore SR
is to SN as SL is to SV; conſequently, the rectangle contained by SR and
SV, or its equal, the rectangle contained by SA and SU is equal to that
contained by SN and SL: but SN is neceſſarily greater than SM, in con-
ſequence whereof the rectangle contained by SN and SL, or its equal, the
rectangle contained by AS and SU is greater than that contained by SM
and SL; wherefore the ratio which the rectangle AS, SU bears to the rect-
angle ES, SI is greater than that which the rectangle SM, SL bears to it,
and of courſe, greater than the ratio which the rectangle AO, UO bears to
the rectangle EO, IO; and that, on which ſide ſoever of the point O, S is
taken.
Again, on YO produced, let fall the perpendiculars EB and IC:
the
triangles EBV and ICY, EBO and ICO are ſimilar, becauſe the angles EVO
and IYO are equal, and ſo EO is to IO as EB is to IC, alſo EV is to IY
as EB is to IC; therefore by equality of ratios EO is to IO as EV is to IY,
and (Eu. VI. 22.) the ſquare on EO is to the ſquare on IO as the ſquare
on EV is to the ſquare on IY; that is (Eu. III. 36.) as the rectangle con-
tained by AE and UE is to that contained by AI and UI.
triangles EBV and ICY, EBO and ICO are ſimilar, becauſe the angles EVO
and IYO are equal, and ſo EO is to IO as EB is to IC, alſo EV is to IY
as EB is to IC; therefore by equality of ratios EO is to IO as EV is to IY,
and (Eu. VI. 22.) the ſquare on EO is to the ſquare on IO as the ſquare
on EV is to the ſquare on IY; that is (Eu. III. 36.) as the rectangle con-
tained by AE and UE is to that contained by AI and UI.
Q.
E.
D.
LEMMA III.
If from two points E and I, in the diameter AU, of a circle, AVYU
(Fig. 30.) two perpendiculars EV, IY be drawn on the ſame ſide thereof to
33[Handwritten note 3] terminate in the periphery, and if their extremes V and Y be joined by a
ſtraight line VY, cutting the ſaid diameter, produced, in O; then will the
ratio which the rectangle contained by AO and UO bears to the rectangle
contained by EO and IO be the greateſt poſſible.
(Fig. 30.) two perpendiculars EV, IY be drawn on the ſame ſide thereof to
33[Handwritten note 3] terminate in the periphery, and if their extremes V and Y be joined by a
ſtraight line VY, cutting the ſaid diameter, produced, in O; then will the
ratio which the rectangle contained by AO and UO bears to the rectangle
contained by EO and IO be the greateſt poſſible.
*** This, like the ſirſt, is demonſtrated by Snellius, and needs not be
repeated.
repeated.