Proof: By Euclid

For let $EH$ and $EK$ have been drawn from the center $E$, at right angles to $BC$ and $FG$ (respectively) [Prop. 1.12].
And since $BC$ is nearer to the center, and $FG$ further away, $EK$ (is) thus greater than $EH$ [Def. 3.5] .
Let $EL$ be made equal to $EH$ [Prop. 1.3].
And $LM$ being drawn through $L$, at right angles to $EK$ [Prop. 1.11], let it have been drawn through to $N$.
And let $ME$, $EN$, $FE$, and $EG$ have been joined.
And since $EH$ is equal to $EL$, $BC$ is also equal to $MN$ [Prop. 3.14].
Again, since $AE$ is equal to $EM$, and $ED$ to $EN$, $AD$ is thus equal to $ME$ and $EN$.
But, $ME$ and $EN$ is greater than $MN$ [Prop. 1.20] [also $AD$ is greater than $MN$], and $MN$ (is) equal to $BC$.
Thus, $AD$ is greater than $BC$.
And since the two (straight lines) $ME$, $EN$ are equal to the two (straight lines) $FE$, $EG$ (respectively), and angle $MEN$ [is] greater than angle $FEG$,¹ the base $MN$ is thus greater than the base $FG$ [Prop. 1.24].
But, $MN$ was shown (to be) equal to $BC$ [(so) $BC$ is also greater than $FG$].
Thus, the diameter $AD$ (is) the greatest (straight line), and $BC$ (is) greater than $FG$.
Thus, in a circle, a diameter (is) the greatest (straight line), and for the others, a (straight line) nearer to the center is always greater than one further away.
(Which is) the very thing it was required to show.

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