Proof: By Euclid
(related to Proposition: 3.22: Opposite Angles of Cyclic Quadrilateral)
 Let $AC$ and $BD$ have been joined.
 Therefore, since the three angles of any triangle are equal to two right angles [Prop. 1.32], the three angles $CAB$, $ABC$, and $BCA$ of triangle $ABC$ are thus equal to two right angles.
 And $CAB$ (is) equal to $BDC$.
 For they are in the same segment $BADC$ [Prop. 3.21].
 And $ACB$ (is equal) to $ADB$.
 For they are in the same segment $ADCB$ [Prop. 3.21].
 Thus, the whole of $ADC$ is equal to $BAC$ and $ACB$.
 Let $ABC$ have been added to both.
 Thus, $ABC$, $BAC$, and $ACB$ are equal to $ABC$ and $ADC$.
 But, $ABC$, $BAC$, and $ACB$ are equal to two right angles.
 Thus, $ABC$ and $ADC$ are also equal to two right angles.
 Similarly, we can show that angles $BAD$ and $DCB$ are also equal to two right angles.
 Thus, for quadrilaterals within circles, the (sum of the) opposite angles is equal to two right angles.
 (Which is) the very thing it was required to show.^{1}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes