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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?

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Show that one vertex lies on the circle determined by the other three. If you don't already "know" that a converse of the Inscribed Angle Theorem (rather, one of its immediate corollaries) gives this to you, then, sure, you could use contradiction effectively to prove the needed converse. But it's also possible to find a related direct approach. – Blue Feb 29 '12 at 14:46
Can anyone prove this by construction? That is, construct a quadrilateral whose opposite angles are supplementary, then prove it is cyclic? – chharvey Sep 2 '12 at 14:22

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.)

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That's essentially the argument I had in mind. For a contradiction-free modification, simply don't make any assumptions about the location of $D$ relative to $ABC$'s circle. Let $E$ be the point where ray $CD$ meets the circle, as above. Considering quads $ABCD$ and $ABCE$, the respective angles at $B$ match, as do the angles at $C$; further, $\angle D$ matches $\angle E$ as supplements to $\angle B$ ($\angle D$ by assumption, $\angle E$ by cyclic quadrilaterality). The quads are thus similar (AAA). As they share edge $BC$, they are in fact congruent. Therefore, points $D$ & $E$ coincide. – Blue Mar 1 '12 at 1:34
A nearly-identical argument proves a more basic fact: for segment $PQ$, the locus of pts $R$ such that $\angle PRQ$ has a particular measure is an arc $PQ$ of a unique circle; necessarily, the other arc is the locus for the supplementary measure. (Right angles give double-arcs.) In the proof, we invoke a corollary of the Inscribed Angle Theorem instead of cyclic quadrilaterality to establish the key similarity (for triangles instead of quads). As I suggested, knowing this solves OP's problem easily: given diagonal $AC$, pt $B$ determines an arc of a circle, and $D$ must lie on the other arc. – Blue Mar 1 '12 at 2:22
@Blue D could be a point on the tangent to the circle at C, in which case the point E would not exist. – user74973 Nov 6 '15 at 0:32

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 $P \neq O$.

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I'm not sure how I would go about doing that... – chharvey Mar 1 '12 at 0:03
If $∠BOC+∠BPC=2π$. then why wouldn't this be possible for P≠O. – user103816 Jul 20 '15 at 10:02

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