117[40]
rectangle contained by AB and AC;
but the ſquare on HO is equal to
the rectangle contained by AB and EC: now EC is, by ſuppoſition,
greater than AC, therefore the rectangle AB, EC is greater than the
rectangle AB, AC, and the ſquare on HO greater than the ſquare on AG,
conſequently HO is itſelf greater than AG; but this could not be the
Caſe unleſs O fell beyond A. In the ſame manner my it be proved that O
will fall beyond U in Fig. 59 and 60.
the rectangle contained by AB and EC: now EC is, by ſuppoſition,
greater than AC, therefore the rectangle AB, EC is greater than the
rectangle AB, AC, and the ſquare on HO greater than the ſquare on AG,
conſequently HO is itſelf greater than AG; but this could not be the
Caſe unleſs O fell beyond A. In the ſame manner my it be proved that O
will fall beyond U in Fig. 59 and 60.
Limitation.
In the above four Caſes the given ratio of R to S muſt
not exceed that which the ſquare on AU bears to the ſquare on the ſum of
two mean proportionals between AI and UE, AE and UI. For (Fig. 30.)
demit from A, on KO produced, the perpendicular AH. Now it has been
proved (Lem. III.) that the ratio of the rectangle continued by AO and UO
to that contained by EO and IO, or which is the ſame thing, the given
ratio of R to S is the greateſt poſſible; and (Lem. IV.) that KF is a mean
proportional between AI and UE, alſo that YF is a mean proportional
between AE and UI: but HK is equal to YF, therefore HF is equal to
the ſum of two mean proportionals between AI and UE, AE and UI;
it only then remains to prove, that the rectangle contained by AO and UO
is to that contained by EO and IO as the ſquare on AU is to the ſquare
on HF. The triangles OEK, OHA, OIY and OUF are all ſimilar; con-
ſequently OK is to OE as OA is to OH, as OY is to OI, and therefore
by compound ratio, the rectangle contained by AO and UO (OK and
OY) is to that contained by EO and IO as the ſquare on AO is to the
ſquare on OH; but alſo AO is to UO as HO is to EO, and by compoſi-
tion and permutation, AU is to HF as AO is to HO, or (Eu. VI. 22.)
the ſquare on AU is to the ſquare on HF as the ſquare on AO is to the
ſquare on HO, and ſo by equality of ratios, the rectangle contained by AO
and UO is to that contained by EO and IO as the ſquare on AU is to the
ſquare on HF.
not exceed that which the ſquare on AU bears to the ſquare on the ſum of
two mean proportionals between AI and UE, AE and UI. For (Fig. 30.)
demit from A, on KO produced, the perpendicular AH. Now it has been
proved (Lem. III.) that the ratio of the rectangle continued by AO and UO
to that contained by EO and IO, or which is the ſame thing, the given
ratio of R to S is the greateſt poſſible; and (Lem. IV.) that KF is a mean
proportional between AI and UE, alſo that YF is a mean proportional
between AE and UI: but HK is equal to YF, therefore HF is equal to
the ſum of two mean proportionals between AI and UE, AE and UI;
it only then remains to prove, that the rectangle contained by AO and UO
is to that contained by EO and IO as the ſquare on AU is to the ſquare
on HF. The triangles OEK, OHA, OIY and OUF are all ſimilar; con-
ſequently OK is to OE as OA is to OH, as OY is to OI, and therefore
by compound ratio, the rectangle contained by AO and UO (OK and
OY) is to that contained by EO and IO as the ſquare on AO is to the
ſquare on OH; but alſo AO is to UO as HO is to EO, and by compoſi-
tion and permutation, AU is to HF as AO is to HO, or (Eu. VI. 22.)
the ſquare on AU is to the ſquare on HF as the ſquare on AO is to the
ſquare on HO, and ſo by equality of ratios, the rectangle contained by AO
and UO is to that contained by EO and IO as the ſquare on AU is to the
ſquare on HF.
Q.
E.
D.
Scholium.
In the four Caſes wherein the points A and U are means,
the limiting ratio will be a minimum, and the ſame with that which the
ſquare on HF bears to the ſquare on EI.
the limiting ratio will be a minimum, and the ſame with that which the
ſquare on HF bears to the ſquare on EI.
THE END.
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