Apollonius <Pergaeus>; Lawson, John - The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies

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107[30] be perpendicular to AF, and FV to AG; whence it follows that GA is
parallel to XI, and the angle NIU equal to the angle UAG; but UAG is
equal to UGE, which is equal to CNE; wherefore, in the triangles UNI,
UNE, the angles at I and N being equal, and that at U common, they
are ſimilar, and UN is to UI as UE is to UN, conſequently the ſquare on
UN is equal to the rectangle contained by UI and UE. Moreover, ſince
IX paſſes through N, and is perpendicular to FU, by Eu. I. 47, the dif-
ference of the ſquares on IF and IU is equal to the difference of the
ſquares on NF and NU: now the ſquare on IF being equal to the rect-
angle contained by AI and UI, that is (Eu. II. I.) to the rectangle con-
tained by AU and UI together with the ſquare on UI, the difference of
the ſquares on IF and UI, and conſequently the difference of the ſquares on
NF and NU is equal to the rectangle contained by AU and UI; but the
ſquare on NU has been proved equal to the rectangle contained by UI
and UE, therefore the ſquare on NF is equal to the rectangle contained
by EU and IU together with that contained by AU and IU, that is (Eu.
II. 1.) to the rectangle contained by AE and UI; wherefore AE is to NF
as NF is to UI.

Laſtly, for the like reaſons which were urged above, the difference of
the ſquares on NU and NG is equal to the difference of thoſe on GE and
UE: now the ſquare on GE is equal to the rectangle contained by AE
and UE, that is, to the rectangle contained by AU and UE together
with the ſquare on UE; therefore the difference of the ſquares on GE
and UE, or the difference of thoſe on NU and NG, is equal to the rect-
angle contained by AU and UE; but the ſquare on NU is equal to the
rectangle contained by UI and UE, therefore the ſquare on NG is
equal to the rectangle contained by AU and UE together with that
contained by UI and UE; that is, to the rectangle contained by AI and
UE, and ſo AI is to NG as NG is to UE. Now FG is equal to the ſum
of NF and NG; therefore FG is equal to the ſum of two mean propor-
tionals between AE and UI, AI and UE.

Q. E. D.

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