84[7]
PROBLEM III.
To cut a given indefinite right line in one point, ſo that of the three ſeg-
ments intercepted between the ſame, and three points given, the rectangle
under two of them may be to the ſquare of the remaining one in a given ratio.
ments intercepted between the ſame, and three points given, the rectangle
under two of them may be to the ſquare of the remaining one in a given ratio.
In the indefinite line let the three points be A, E, I.
it is then required to be
cut again in O, ſo that OA x OE may be to OI2 (let the ſituation of I be
what it may) in a given ratio, which ratio let be expreſſed by EL to LI.
[And here I cannot but obſerve with Hugo D'Omerique, page 113. that
this Problem, viz. ‘To exhibit two lines in a given ratio whoſe ſum, or whoſe
difference is given,’ ought to have had a place in the Elements as a Propoſition;
or at leaſt to have been annext as a Scholium to the 9th or 10th of the VIth
Book. ] And be the ſituation of L alſo what it may, either between A and E,
or between A and I, or between E and I, or beyond either extreme. To the three
points E, L, I, and the right line AI, let be found, by Problem II, a fourth
point O ſuch, that AI x OE: OI x OL: : EI: IL. And let ſuch a Caſe be
choſen of Problem II, that, according as AO is greater or leſs than AI, ſo of
the three rectangles, deſcribed in Lemma V, made by the four points E, O,
I, L, that of IO x EL may accordingly be greater or leſs than that of EI x OL.
cut again in O, ſo that OA x OE may be to OI2 (let the ſituation of I be
what it may) in a given ratio, which ratio let be expreſſed by EL to LI.
[And here I cannot but obſerve with Hugo D'Omerique, page 113. that
this Problem, viz. ‘To exhibit two lines in a given ratio whoſe ſum, or whoſe
difference is given,’ ought to have had a place in the Elements as a Propoſition;
or at leaſt to have been annext as a Scholium to the 9th or 10th of the VIth
Book. ] And be the ſituation of L alſo what it may, either between A and E,
or between A and I, or between E and I, or beyond either extreme. To the three
points E, L, I, and the right line AI, let be found, by Problem II, a fourth
point O ſuch, that AI x OE: OI x OL: : EI: IL. And let ſuch a Caſe be
choſen of Problem II, that, according as AO is greater or leſs than AI, ſo of
the three rectangles, deſcribed in Lemma V, made by the four points E, O,
I, L, that of IO x EL may accordingly be greater or leſs than that of EI x OL.
DEMONSTRATION .
On ſuppoſition then that ſuch a Caſe of Problem II.
is
made uſe of, we have
AI x OE: OI x OL: : EI: IL
made uſe of, we have
AI x OE: OI x OL: : EI: IL
And by Lemma IV, OL x EI:
OE x IL:
: AI:
OI
And by Diviſion or Compoſition EL x OI:
OE x IL:
: AO:
OI
This appears from Lemma V.
Then again by Lemma IV, AO x OE:
OI2:
: EL:
IL.
Q.
E.
D.
This Problem has two Epitagmas.
The firſt wherein OI, whoſe ſquare is
ſought, is bounded by I an extreme point of the three given ones. And this
again admits of three Caſes. The ſecond is when the point I is the middle
point. And this again has three caſes. And there remain two Anomalous
Caſes, wherein Problem II. is of no uſe, which muſt therefore be conſtructed
by themſelves.
ſought, is bounded by I an extreme point of the three given ones. And this
again admits of three Caſes. The ſecond is when the point I is the middle
point. And this again has three caſes. And there remain two Anomalous
Caſes, wherein Problem II. is of no uſe, which muſt therefore be conſtructed
by themſelves.
Epitagma I.
Case I.
Let the ratio given, EL to LI, be inequalitatis
majoris, i. e. of a greater to a leſs; and the point O ſought be required to lie
between I and the next point to it E, or elſe to lie beyond I the other way;
for the ſame conſtruction ſerves for both. Here Case I. of Problem II. is
majoris, i. e. of a greater to a leſs; and the point O ſought be required to lie
between I and the next point to it E, or elſe to lie beyond I the other way;
for the ſame conſtruction ſerves for both. Here Case I. of Problem II. is