91[14]
Demonstration.
Through S, any other point taken at pleaſure, draw
LSM parallel to VOY, and join VS and produce it to meet the perpendi-
cular in N and the circumference in R. Produce alfo the perpendicular YI
to meet the circumference again in F, and join RF; then from ſrmilar tri-
angles it appears that the rectangle LSM: ESI: : VOY i. e. AOU: EOI.
But the rectangle ASU i. e. VSR is greater than LSM. (for LV i. e. MY x
NI + MI together with VS x SN is by the preceeding Lemma equal to LSM.
But ASU or VSR is equal to VS x SN, together with VS x NR; for which
laſt rectangle we may ſubſtitute MY x NF: for the triangles VLS and NRF
are ſimilar, being each of them ſimilar to VNY; therefore VL or MY: VS
: : NR: NF and MY x NF = VS x NR. Now NF being always greater
than NI + MI, it appears from thence that ASU is greater than LSM.)
Therefore the ratio of ASU to ESI is greater than that of AOU to EOI.
And the ſame holds good with regard to any other point S taken between E
and I, ſo that the ratio of AOU to EOI is a Minimum, and ſingular, or
what the Antients called μοναχ@.
LSM parallel to VOY, and join VS and produce it to meet the perpendi-
cular in N and the circumference in R. Produce alfo the perpendicular YI
to meet the circumference again in F, and join RF; then from ſrmilar tri-
angles it appears that the rectangle LSM: ESI: : VOY i. e. AOU: EOI.
But the rectangle ASU i. e. VSR is greater than LSM. (for LV i. e. MY x
NI + MI together with VS x SN is by the preceeding Lemma equal to LSM.
But ASU or VSR is equal to VS x SN, together with VS x NR; for which
laſt rectangle we may ſubſtitute MY x NF: for the triangles VLS and NRF
are ſimilar, being each of them ſimilar to VNY; therefore VL or MY: VS
: : NR: NF and MY x NF = VS x NR. Now NF being always greater
than NI + MI, it appears from thence that ASU is greater than LSM.)
Therefore the ratio of ASU to ESI is greater than that of AOU to EOI.
And the ſame holds good with regard to any other point S taken between E
and I, ſo that the ratio of AOU to EOI is a Minimum, and ſingular, or
what the Antients called μοναχ@.
Let now VY join the tops of two perpendiculars drawn on the ſame ſide of
the diameter, and meet the diameter produced in O; I ſay that the ratio of
AOU to EOI is a Maximum.
the diameter, and meet the diameter produced in O; I ſay that the ratio of
AOU to EOI is a Maximum.
For uſing the ſame conſtruction as before, it will appear that the rectangle
LSM: ESI: : VOY or AOU: EOI. And it may be proved in the ſame
manner that VSR or ASU is leſs than LSM. (LV i. e. MY x NI + MI or
(making IK = MI) MY x NK, together with LSM is by the preceding Lem-
MA equal to VS x SN. But VS x NR, together with VSR, is alſo equal to
VS x SN. Now VS x NR is equal to MY x NF or LV x NF from the ſimi-
larity of the triangles LVS, NRF. Therefore now alſo MY x NF together
with VSR is proved equal to VS x SN. But as NK is leſs than NF, VSR
will be a leſs rectangle than LSM) Hence the ratio of LSM to ESI or it's
equal AOU to EOI is greater than the ratio of VSR or ASU to ESI. And
the ſame holds with regard to any other point taken in the diameter pro-
duced. Therefore the ratio of AOU to EOI is a Maximum.
LSM: ESI: : VOY or AOU: EOI. And it may be proved in the ſame
manner that VSR or ASU is leſs than LSM. (LV i. e. MY x NI + MI or
(making IK = MI) MY x NK, together with LSM is by the preceding Lem-
MA equal to VS x SN. But VS x NR, together with VSR, is alſo equal to
VS x SN. Now VS x NR is equal to MY x NF or LV x NF from the ſimi-
larity of the triangles LVS, NRF. Therefore now alſo MY x NF together
with VSR is proved equal to VS x SN. But as NK is leſs than NF, VSR
will be a leſs rectangle than LSM) Hence the ratio of LSM to ESI or it's
equal AOU to EOI is greater than the ratio of VSR or ASU to ESI. And
the ſame holds with regard to any other point taken in the diameter pro-
duced. Therefore the ratio of AOU to EOI is a Maximum.
Q.
E.
D.