# Proof: By Euclid

• 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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