Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles
Theorem
In the words of Euclid:
- The opposite angles of quadrilaterals in circles are equal to two right angles.
(The Elements: Book $\text{III}$: Proposition $22$)
Proof
Let $ABCD$ be a cyclic quadrilateral in a circle.
Join $AC$ and $BD$.
From Sum of Angles of Triangle equals Two Right Angles, we have that $\angle CAB + \angle ABC + \angle BCA$ equals two right angles.
But from Angles in Same Segment of Circle are Equal we have that $\angle CAB = \angle BDC$ and also $\angle ACB = \angle ADB$.
If we add $\angle ABC$ to each, we get that $\angle ABC + \angle BAC + \angle ACB = \angle ABC + \angle ADC$.
Therefore $\angle ABC + \angle ADC$ equals two right angles.
Similarly we can show that $\angle BAD + \angle DCB$ also equals two right angles.
$\blacksquare$
Historical Note
This proof is Proposition $22$ of Book $\text{III}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): cyclic quadrilateral
- 1991: David Wells: Curious and Interesting Geometry ... (previous) ... (next): angle in the same segment