Proof: By Euclid
(related to Proposition: 4.01: Fitting Chord Into Circle)
 Let a diameter $BC$ of circle $ABC$ have been drawn.^{1} Therefore, if $BC$ is equal to $D$ then that (which) was prescribed has taken place.
 For the (straight line) $BC$, equal to the straight line $D$, has been inserted into the circle $ABC$.
 And if $BC$ is greater than $D$ then let $CE$ be made equal to $D$ [Prop. 1.3], and let the circle $EAF$ have been drawn with center $C$ and radius $CE$.
 And let $CA$ have been joined.
 Therefore, since the point $C$ is the center of circle $EAF$, $CA$ is equal to $CE$.
 But, $CE$ is equal to $D$.
 Thus, $D$ is also equal to $CA$.
 Thus, $CA$, equal to the given straight line $D$, has been inserted into the given circle $ABC$.
 (Which is) the very thing it was required to do.^{2}
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes