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This is a proof that the sum of the measures of opposite angles in any simple cyclic quadrilateral is always $180^\circ$.

Let the polygon with vertices $A$, $B$, $C$ and $D$ be a simple cyclic quadrilateral. Next, construct one of the quadrilateral's diagonals (for explanatory purposes, make it AC). Then the sum of the arcs subtended by angles $B$ and $D$ will be (by definition) equal to the circumference of the circumscribed circle ($360^\circ$). So the sum of angles $B$ and $D$ is $180^\circ$.

I was told that this proof isn't complete; that something is wrong with it. But I don't see what. What is wrong with this proof?

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    Your proof is correct.2011-08-01

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I agree with Srivatsan - if we know that the angle subtended by an angle with a vertex on the circle is half the angle from the center of the circle, then it seems to me that OP's original argument is true. As I recall the original proof of this theorem,there are 3 cases: (1) one line of the angle passes through the center (and we use the theorem that an exterior angle is the sum of the opposite interior angles); (2) the angle contains the center; (3) the angle does not contain the center.