Prove the opposite angles of a quadrilateral are supplementary implies it is cyclic. There is a well-known theorem that a cyclic quadrilateral (its vertices all lie on the same circle) has supplementary opposite angles.
I have a feeling the converse is true, but I don't know how to prove it. The converse states:

If a quadrilateral's opposite angles are supplementary then it is cyclic.

Should I approach this proof by contradiction? Or is it possible to prove by construction?
 A: A proof by contradiction is a good approach. Suppose you have a quadrilateral $ABCD$ whose opposite angles are supplementary, but it is not cyclic. The vertices $A,B,C$ determine a circle, and the point $D$ does not lie on this circle, since we assume the quadrilateral is not cyclic.
Suppose for instance that $D$ lies outside the circle, and so the circle intersects $ABCD$ at some point $E$ on $CD$ (try drawing a picture to see this if needed.) Now $D$ is supplementary to $B$, and since $E$ is the opposite angle of $B$ in the cyclic quadrilateral $ABCE$, $E$ is supplementary to $B$ by the theorem you already know, and so $D$ and $E$ are congruent. But this contradicts the fact that an exterior angle cannot be congruent to an interior angle, which proves the converse. A similar method works if $D$ lies inside the circle as well. (I abuse notation a bit and refer to a vertex and the angle at that vertex by the same letter.)
A: Denote $ABCD$ your quadrilateral. You can prove that $ABC$ and $BCD$ have the same circumcenter.

A short argument.
Denote by $O$ the circumcenter of $ABC$ and $P$ the circumcenter of $BCD$. Then both triangles $BOC$ and $BPC$ are isosceles, $\angle BOC=2\angle A, \angle BPC=2\angle D$. Therefore $\angle BOC+\angle BPC=2\pi$. This wouldn't be possible if $P \neq O$.
