Aha... Actually it's easy to prove. Let the four corners of the cyclic quad be $A$, $B$, $C$, $D$ (in clockwise order) and the centre of the circle by $O$. Now $OA=OB=OC=OD$ so the triangles $OAB$, $OBC$, $OCD$, $ODA$ are isosceles. Therefore:

angle $OAB$ = angle $OBA$ [1]

angle $OBC$ = angle $OCB$ [2]

angle $OCD$ = angle $ODC$ [3]

angle $ODA$ = angle $OAD$ [4]

Define the angles on each side of equations [1]-[4] to be $p$, $q$, $r$, $s$ respectively. Now using the fact that angles in a triangle add to $\pi$, we get:

angle $AOB=\pi -2p$

angle $BOC=\pi -2q$

angle $COD=\pi -2r$

angle $DOA=\pi -2s$

These four angles must add to $2\pi$. So:

$2\pi =4\pi -2(p+q+r+s)$

$p+q+r+s=\pi$

So angle $ADC$ + angle $ABC$ = $\pi$.

So the two opposite angles in a cyclic quad add to $\pi$.