103[26]
DETERMINATE SECTION.
BOOK II.
LEMMA I.
BOOK II.
LEMMA I.
If from two points E and I in the diameter AU of a circle AYUV (Fig.
27.) two perpendiculars EV, IY be drawn contrary ways to terminate in
the Circumference; and if their extremes V and Y be joined by a ſtraight
line VY, cutting the faid diameter 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 leaſt poſſible.
27.) two perpendiculars EV, IY be drawn contrary ways to terminate in
the Circumference; and if their extremes V and Y be joined by a ſtraight
line VY, cutting the faid diameter 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 leaſt poſſible.
*** This being demonſtrated in the preceeding Tract of Snellius, I
ſhall not attempt it here.
ſhall not attempt it here.
LEMMA II.
If to a circle deſcribed on AU, tangents EV, IY (Fig.
28.
29.)
be drawn
from E and I, two points in the diameter AU produced, and through the
points of contact V, and Y, a ſtraight line YVO be drawn to cut the line
AI in O; then will the ratio which the rectangle contained by AO and
UO bears to that contained by EO and IO be the leaſt poſſible: and
moreover, the ſquare on EO will be the ſquare on IO as the rectangle con-
tained by AE and UE is to the rectangle contained by AI and UI.
from E and I, two points in the diameter AU produced, and through the
points of contact V, and Y, a ſtraight line YVO be drawn to cut the line
AI in O; then will the ratio which the rectangle contained by AO and
UO bears to that contained by EO and IO be the leaſt poſſible: and
moreover, the ſquare on EO will be the ſquare on IO as the rectangle con-
tained by AE and UE is to the rectangle contained by AI and UI.
Demonstration.
If the ſaid ratio be not then a minimum, let it be
when the ſegments are bounded by ſome other point S, through which and
the point V, let the ſtraight line SV be drawn, meeting the circle again in
R; draw SM parallel to OY, meeting the tangents EV and IY in L and
M, and through R and Y draw the ſtraight line RY meeting SM produced
in N: the triangles ESL and EOV, ISM and IOY are ſimilar; wherefore
LS is to SE as VO is to EO, and SM is to SI as YO is to OI;
when the ſegments are bounded by ſome other point S, through which and
the point V, let the ſtraight line SV be drawn, meeting the circle again in
R; draw SM parallel to OY, meeting the tangents EV and IY in L and
M, and through R and Y draw the ſtraight line RY meeting SM produced
in N: the triangles ESL and EOV, ISM and IOY are ſimilar; wherefore
LS is to SE as VO is to EO, and SM is to SI as YO is to OI;