105[28]
LEMMA IV.
If EK and IY (Fig.
27.)
be any perpendiculars to the diameter AU of
a circle AYUV, terminating in the circumference, and if KY be drawn,
on which, from U, the perpendicular UF is demitted; then will KF be a
mean proportional between AI and EU, alſo YF a mean proportional
between AE and IU.
a circle AYUV, terminating in the circumference, and if KY be drawn,
on which, from U, the perpendicular UF is demitted; then will KF be a
mean proportional between AI and EU, alſo YF a mean proportional
between AE and IU.
Demonstration.
Draw UY, UK, KA and AY, the angles I and F
44[Handwritten note 4]55[Handwritten note 5] being right by conſtruction, and the angles IDU, and FKV equal, being
44[Handwritten note 4]55[Handwritten note 5] both equal to the angle UAY, the triangles IYU and FKU are ſimilar,
and conſequently IY is to UY as KF is to UK; or (Eu. VI. 22.) the
ſquare on IY is to the ſquare on UY as the ſquare on KF is to the ſquare
on UK: now the ſquare on IY is (Eu. VI. 8. 17.) equal to the rectangle
contained by AI and IU, the ſquare on UY to the rectangle contained by
AU and IU, and the ſquare on UK to the rectangle contained by AU and
EU; wherefore the rectangle contained by AI and IU is to that contained
by AU and IU as the ſquare on KF is to the rectangle contained by AU
and UE, whence (Eu. V. 15.) AI is to AU as the ſquare on KF is to the
rectangle contained by AU and UE, or the rectangle contained by AI and
UE is to that contained by AU and UE as the ſquare on KF is to the
rectangle contained by AU and UE; ſeeing then that the conſequents are
here the ſame, the antecedents muſt be equal, and therefore AI is to KF
as KF is to UE.
44[Handwritten note 4]55[Handwritten note 5] being right by conſtruction, and the angles IDU, and FKV equal, being
44[Handwritten note 4]55[Handwritten note 5] both equal to the angle UAY, the triangles IYU and FKU are ſimilar,
and conſequently IY is to UY as KF is to UK; or (Eu. VI. 22.) the
ſquare on IY is to the ſquare on UY as the ſquare on KF is to the ſquare
on UK: now the ſquare on IY is (Eu. VI. 8. 17.) equal to the rectangle
contained by AI and IU, the ſquare on UY to the rectangle contained by
AU and IU, and the ſquare on UK to the rectangle contained by AU and
EU; wherefore the rectangle contained by AI and IU is to that contained
by AU and IU as the ſquare on KF is to the rectangle contained by AU
and UE, whence (Eu. V. 15.) AI is to AU as the ſquare on KF is to the
rectangle contained by AU and UE, or the rectangle contained by AI and
UE is to that contained by AU and UE as the ſquare on KF is to the
rectangle contained by AU and UE; ſeeing then that the conſequents are
here the ſame, the antecedents muſt be equal, and therefore AI is to KF
as KF is to UE.
Again, the angle AKE is equal to AUK, which is equal to the angle
AYK, of which the angle UYF is the complement, becauſe AYU is a
right angle; and therefore as the angles F and E are both right, the tri-
angles AKE and YUF are ſimilar, and AK is to AE as YU is to YF,
wherefore the ſquare on AK is to the ſquare on AE as the ſquare on YU
is to the ſquare on YF: but the ſquare on AK is equal to the rectangle
contained by AU and AE, and the ſquare on YU is equal to the rectangle
contained by AU and IU, conſequently the rectangle contained by AU
and AE is to the ſquare on AE as the rectangle contained by AU and IU
is to the ſquare on YF; whence (Eu. V. 15.) AU is to AE as the rect-
angle contained by AU and IU is to the ſquare on YF, or the rectangle
contained by AU and IU is to that contained by AE and IU as the
AYK, of which the angle UYF is the complement, becauſe AYU is a
right angle; and therefore as the angles F and E are both right, the tri-
angles AKE and YUF are ſimilar, and AK is to AE as YU is to YF,
wherefore the ſquare on AK is to the ſquare on AE as the ſquare on YU
is to the ſquare on YF: but the ſquare on AK is equal to the rectangle
contained by AU and AE, and the ſquare on YU is equal to the rectangle
contained by AU and IU, conſequently the rectangle contained by AU
and AE is to the ſquare on AE as the rectangle contained by AU and IU
is to the ſquare on YF; whence (Eu. V. 15.) AU is to AE as the rect-
angle contained by AU and IU is to the ſquare on YF, or the rectangle
contained by AU and IU is to that contained by AE and IU as the
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