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Case VI.
Let the given ratio of EL to LI be inæqualitatis majoris, and
let the point ſought be required to lie beyond either extreme. Here we muſt
uſe the IIId Case of the IId Problem; and the Determination is that
UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than
IE + EL + √4 IEL*.
let the point ſought be required to lie beyond either extreme. Here we muſt
uſe the IIId Case of the IId Problem; and the Determination is that
UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than
IE + EL + √4 IEL*.
Case VII.
Let the ſituation of O be required the ſame as in the two laſt
Caſes, but let the given ratio be that of equality, which was there ſuppoſed
of inequality. Here the IId Problem will be of no uſe, and this Caſe
requires a particular conſtruction.
Caſes, but let the given ratio be that of equality, which was there ſuppoſed
of inequality. Here the IId Problem will be of no uſe, and this Caſe
requires a particular conſtruction.
Let then the three Points be A, I, E, and I the middle one;
and let it
be required to find a fourth O beyond E, ſuch that AO x OE may equal
OI2.
be required to find a fourth O beyond E, ſuch that AO x OE may equal
OI2.
Construction.
Upon AE diameter deſcribe a circle,, and let another YS
cut the former at right angles. Join SI, and continue it to meet the circum-
ference in R. From R draw a tangent to meet the given line in O, and I
ſay O is the point required.
cut the former at right angles. Join SI, and continue it to meet the circum-
ference in R. From R draw a tangent to meet the given line in O, and I
ſay O is the point required.
Demonstration.
Joining YR, the triangles SUI and SYR will be ſimi-
lar, and the angle UIS or RIO = SYR. But the angle IRO made by the
tangent and ſecant = SYR in the alternate ſegment. Therefore RIO = IRO,
and OR = OI. But by the property of the circle AOE = OR2. And
therefore AOE = OI2.
lar, and the angle UIS or RIO = SYR. But the angle IRO made by the
tangent and ſecant = SYR in the alternate ſegment. Therefore RIO = IRO,
and OR = OI. But by the property of the circle AOE = OR2. And
therefore AOE = OI2.