Proof: By Euclid
(related to Lemma: Lem. 12.02: Areas of Circles are as Squares on Diameters)
 For let it have been contrived that as area $S$ (is) to circle $ABCD$, so circle $EFGH$ (is) to area $T$.
 I say that area $T$ is less than circle $ABCD$.
 For since as area $S$ is to circle $ABCD$, so circle $EFGH$ (is) to area $T$, alternately, as area $S$ is to circle $EFGH$, so circle $ABCD$ (is) to area $T$ [Prop. 5.16].
 And area $S$ (is) greater than circle $EFGH$.
 Thus, circle $ABCD$ (is) also greater than area $T$ [Prop. 5.14].
 Hence, as area $S$ is to circle $ABCD$, so circle $EFGH$ (is) to some area less than circle $ABCD$.
 (Which is) the very thing it was required to show.
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"