Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles
In the words of Euclid:
Join $AC$ and $BD$.
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.