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Proof: By Euclid

(related to Proposition: 3.34: Construction of Segment on Given Circle Admitting Given Angle)

• Let $EF$ have been drawn touching $ABC$ at point $B$.¹ And Let (angle) $FBC$, equal to angle $D$, have been constructed on the straight line $FB$, at the point $B$ on it [Prop. 1.23].
• Therefore, since some straight line $EF$ touches the circle $ABC$, and $BC$ has been drawn across (the circle) from the point of contact $B$, angle $FBC$ is thus equal to the angle constructed in the alternate segment $BAC$ [Prop. 1.32].
• But, $FBC$ is equal to $D$.
• Thus, the (angle) in the segment $BAC$ is also equal to [angle] $D$.
• Thus, the segment $BAC$, accepting an angle equal to the given rectilinear angle $D$, has been cut off from 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)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"

Footnotes