90[13]
ſuppoſe it that of equality, and they may all be ſolved by one conſtruction, viz.
the rectangle AOU made equal to the rectangle EOI.
the rectangle AOU made equal to the rectangle EOI.
Hence AOU = EOI.
Lemma VI.
Let there be two ſimilar triangles IAE, UAO, having their
baſes IE and UO parallel; I ſay Iſt when they are right-angled, that the ex-
ceſs of the rectangle EAO, under the greater ſides of each, above the rect-
angle IAU, under the leſſer ſides of each, will be equal to the rectangle
IE x OU, under their baſes. IIdly, When they are obtuſe-angled, that the
ſaid exceſs will be equal to the rectangle under the baſe of one and the ſum
of the diſtances of the angles at the baſe of the other from the perpendicular,
viz. EI x OS + US. IIIdly, When they are acute-angled, that then the ſaid
exceſs will be equal to the rectangle under the baſe of one and the difference
of the ſegments of the baſe of the other made by the perpendicular, viz.
OU x EL.
baſes IE and UO parallel; I ſay Iſt when they are right-angled, that the ex-
ceſs of the rectangle EAO, under the greater ſides of each, above the rect-
angle IAU, under the leſſer ſides of each, will be equal to the rectangle
IE x OU, under their baſes. IIdly, When they are obtuſe-angled, that the
ſaid exceſs will be equal to the rectangle under the baſe of one and the ſum
of the diſtances of the angles at the baſe of the other from the perpendicular,
viz. EI x OS + US. IIIdly, When they are acute-angled, that then the ſaid
exceſs will be equal to the rectangle under the baſe of one and the difference
of the ſegments of the baſe of the other made by the perpendicular, viz.
OU x EL.
Demonstration.
Since EA:
AO:
: IA:
AU:
: EI:
OU, the rect-
angles EAO, IAU, and EI x OU will be ſimilar, and when Iſt the triangles
are right-angled EAO = IAU + EI x OU by Euc. VI. 19. and I. 47. But if
they be oblique-angled, draw the perpendicular YAS. Then IIdly, in caſe
they be obtuſe-angled, EAO = YAS + EY x OS by part Iſt; and IAU =
YAS + IY x US by the ſame. And therefore EAO - IAU = EY x OS -
IY x US = EY - IY or EI x OS + US. But if IIIdly they be acute-angled,
and EY be greater than IY, then from Y ſet off YL = YI, and draw LAR
which will be equal and ſimilarly divided to IAU. Then by part IId EAO
- LAR, i. e. EAO - IAU = EL x OS + RS = EL x OU.
angles EAO, IAU, and EI x OU will be ſimilar, and when Iſt the triangles
are right-angled EAO = IAU + EI x OU by Euc. VI. 19. and I. 47. But if
they be oblique-angled, draw the perpendicular YAS. Then IIdly, in caſe
they be obtuſe-angled, EAO = YAS + EY x OS by part Iſt; and IAU =
YAS + IY x US by the ſame. And therefore EAO - IAU = EY x OS -
IY x US = EY - IY or EI x OS + US. But if IIIdly they be acute-angled,
and EY be greater than IY, then from Y ſet off YL = YI, and draw LAR
which will be equal and ſimilarly divided to IAU. Then by part IId EAO
- LAR, i. e. EAO - IAU = EL x OS + RS = EL x OU.
Lemma VII.
If a right line VY, joining the tops of two perpendiculars
drawn from two points of the diameter of a circle E and I to the circum-
ference on oppoſite ſides of the diameter, cut the ſaid diameter in O, and
A and U be the extremes of the ſaid diameter, I ſay that the ratio of the
rectangle AOU to the rectangle EOI is a Minimum.
drawn from two points of the diameter of a circle E and I to the circum-
ference on oppoſite ſides of the diameter, cut the ſaid diameter in O, and
A and U be the extremes of the ſaid diameter, I ſay that the ratio of the
rectangle AOU to the rectangle EOI is a Minimum.