Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles

Proof

 * Euclid-III-22.png

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 $\angle CAB + \angle ABC + \angle CBA$ 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.