79[2]
but if it only touches, then only of one;
if it neither touches nor cuts, it is
then impoſſible.
then impoſſible.
Demonstration.
Let a point of interſection then be O, and join O Y
and OR. The angles AYO and IOR are equal, the angle AOY being the
complement of each of them to a right one, and hence the triangles AOY and
IOR are ſimilar.
and OR. The angles AYO and IOR are equal, the angle AOY being the
complement of each of them to a right one, and hence the triangles AOY and
IOR are ſimilar.
Hence AY = AE:
AO:
: OI:
IR = AI
And by div . or comp . EO: AO: : AO: AI
And AO2 = EO x AI
Therefore AO2 (= EO x AI): EO x AU: : AI: AU: : R: S
Q. E. D.
And by div . or comp . EO: AO: : AO: AI
And AO2 = EO x AI
Therefore AO2 (= EO x AI): EO x AU: : AI: AU: : R: S
Q. E. D.
This Problem admits of two Caſes.
The 1ſt determinate or limited, the 2d
unlimited.
unlimited.
Case I.
Is when OE the co-efficient of the given external line AU is part of
AO the ſide of the required ſquare [fig. 1. 2. ] and here the Llmitation is,
that AI muſt not be given leſs than four times AE, as appears from fig. 2.
for AE: AO: : OI: AI; and here OI being the half of AI, AE will be
the half of AO, or the fourth part of AI. In this Caſe the Homotactical Con-
ſtruction is uſed.
AO the ſide of the required ſquare [fig. 1. 2. ] and here the Llmitation is,
that AI muſt not be given leſs than four times AE, as appears from fig. 2.
for AE: AO: : OI: AI; and here OI being the half of AI, AE will be
the half of AO, or the fourth part of AI. In this Caſe the Homotactical Con-
ſtruction is uſed.
Case II.
Is when AO the ſide of the required ſquare is part of OE the co-
efficient of the given external line AU, [fig. 3. ] and this is unlimited, for here
the Anlitactical Conſtruction is uſed. Or if O be required between A and E,
this is effected by the ſame Conſtruction.
efficient of the given external line AU, [fig. 3. ] and this is unlimited, for here
the Anlitactical Conſtruction is uſed. Or if O be required between A and E,
this is effected by the ſame Conſtruction.
LEMMA I.
If from the extremes of any diameter perpendiculars be let fall upon any
Chord, I ſay that the ſegments of theſe perpendiculars intercepted by like Arcs
are equal, and moreover alſo the ſegments of the Chords themſelves.
Chord, I ſay that the ſegments of theſe perpendiculars intercepted by like Arcs
are equal, and moreover alſo the ſegments of the Chords themſelves.
That YO is equal to IU may be thus ſhewn.
Having joined YI, the
angle IYE is a right one, being in a ſemicircle, and the angle at O is right by
hypotbeſis; hence YI is parallel to the Chord, and YOUI is a parallelogram,
and the oppoſite ſides YO and IU will be equal. In the ſame manner OE is
proved equal to UL. And as to the ſegments of the Chord, it is thus ſhewn.
By Euc. III. 35. and 36, the rect. EOY = rect. SOR, and rect. LUI = rect.
SUR. But, by what has been juſt proved, rect. EOY = rect. LUI; hence
rect. SOR = rect. SUR, and the ſegments SO and OR are reſpectively equal
to the ſegments UR and SU.
angle IYE is a right one, being in a ſemicircle, and the angle at O is right by
hypotbeſis; hence YI is parallel to the Chord, and YOUI is a parallelogram,
and the oppoſite ſides YO and IU will be equal. In the ſame manner OE is
proved equal to UL. And as to the ſegments of the Chord, it is thus ſhewn.
By Euc. III. 35. and 36, the rect. EOY = rect. SOR, and rect. LUI = rect.
SUR. But, by what has been juſt proved, rect. EOY = rect. LUI; hence
rect. SOR = rect. SUR, and the ſegments SO and OR are reſpectively equal
to the ſegments UR and SU.
zoom in
zoom out
zoom area
full page
page width
set mark
remove mark
get reference
digilib