If in any ſtraight line four points A, U, E and I (Fig.
31.)
be aſſigned,
and if the point O be ſo taken by Lemma II, that the ratio of the rect-
angle contained by AO and UO to that contained by EO and IO may be
the leaſt poſſible; alſo if through O the indefinite perpendicular FG be
drawn; and laſtly, if from E and I, EG and IF be applied to FG, the
former equal to a mean proportional between AE and UE, and the latter
to one between AI and UI: then ſhall FG be equal to the ſum of two
mean proportionals between AE and UI, AI and UE.
and if the point O be ſo taken by Lemma II, that the ratio of the rect-
angle contained by AO and UO to that contained by EO and IO may be
the leaſt poſſible; alſo if through O the indefinite perpendicular FG be
drawn; and laſtly, if from E and I, EG and IF be applied to FG, the
former equal to a mean proportional between AE and UE, and the latter
to one between AI and UI: then ſhall FG be equal to the ſum of two
mean proportionals between AE and UI, AI and UE.
Demonstration.
Draw AF and AG, and, through U, FV and GY,
produce GE to meet FV in H, and let fall on FV the perpendicular XI,
cutting FG in N; moreover draw UM through N, and NP through E,
and theſe two laſt will be reſpectively perpendiculars to IF and UG, be-
cauſe the three perpendiculars of every plane triangle meet in a point.
Since by conſtruction and Eu. VI. 17, the ſquare on EG is equal to the
rectangle contained by AE and UE, and the ſquare on IF to that con-
tained by AI and UI, and becauſe (Lem. II.) the ſquare on EO is to the
ſquare on IO as the rectangle AE, UE is to the rectangle AI, UI; the
ſquare on EO is to the ſquare on IO as the ſquare on EG is to the ſquare
on IF, and (Eu. VI. 22.) EO is to IO as EG is to IF; from whence it
appears that the triangles EOG and IOF are ſimilar, and HG parallel to
IF, and the angle UHE equal to the angle UFI. Again, becauſe AI is
to IF as IF is to UI, the triangles AIF, and UFI are ſimilar, and ſo, for
like reaſons, are the triangles AEG and GEU, wherefore the angles UFI
and FAE are equal, and alſo the angles UGE and UAG; hence there-
fore (Eu. I 32.) the angle YUF is equal to the angle UAF (UHE) toge-
ther with the angle UAG (UGE) and conſequently, the angles VAY and
YUV are together equal to two right angles; wherefore the points AYUV
are in a circle: hence, and becauſe AO is perpendicular to FG, GY 1
produce GE to meet FV in H, and let fall on FV the perpendicular XI,
cutting FG in N; moreover draw UM through N, and NP through E,
and theſe two laſt will be reſpectively perpendiculars to IF and UG, be-
cauſe the three perpendiculars of every plane triangle meet in a point.
Since by conſtruction and Eu. VI. 17, the ſquare on EG is equal to the
rectangle contained by AE and UE, and the ſquare on IF to that con-
tained by AI and UI, and becauſe (Lem. II.) the ſquare on EO is to the
ſquare on IO as the rectangle AE, UE is to the rectangle AI, UI; the
ſquare on EO is to the ſquare on IO as the ſquare on EG is to the ſquare
on IF, and (Eu. VI. 22.) EO is to IO as EG is to IF; from whence it
appears that the triangles EOG and IOF are ſimilar, and HG parallel to
IF, and the angle UHE equal to the angle UFI. Again, becauſe AI is
to IF as IF is to UI, the triangles AIF, and UFI are ſimilar, and ſo, for
like reaſons, are the triangles AEG and GEU, wherefore the angles UFI
and FAE are equal, and alſo the angles UGE and UAG; hence there-
fore (Eu. I 32.) the angle YUF is equal to the angle UAF (UHE) toge-
ther with the angle UAG (UGE) and conſequently, the angles VAY and
YUV are together equal to two right angles; wherefore the points AYUV
are in a circle: hence, and becauſe AO is perpendicular to FG, GY 1