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?

  • $\begingroup$ Why the jump to "...so the sum of angles B and D is 180°" from the statement before it? $\endgroup$ – J. M. is a poor mathematician Aug 1 '11 at 2:38
  • $\begingroup$ The entire weight of the proof is on the final implication that if the sum of the arcs subtended by angles $B$ and $D$ is the circumference of the circle, then the sum of the angles is $180^{\circ}$. But I do not see any way of proving this without invoking the fact that opposite angles of a cyclic quadrilateral sum to $180^{\circ}$. $\endgroup$ – Srivatsan Aug 1 '11 at 2:47
  • $\begingroup$ Of course, you can salvage the proof in the following way. Look at the angle subtended by the arcs $ABC$ and $ADC$ at the center; these sum to $360^{\circ}$. Now, the angle $B$ is simply half the angle subtended by the arc $ABC$, and similarly for angle $D$. Therefore, $B+D$ is $180^{\circ}$. Now, it becomes a valid proof. Unfortunately, from what I remember, this was also the proof that I learned at school :-) . $\endgroup$ – Srivatsan Aug 1 '11 at 2:57
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    $\begingroup$ You seem to be using the theorem that the measure of an inscribed angle is half the measure of the intercepted arc. If you cite this theorem then your argument is OK. Your argument makes no use of the segment AC, so I would delete that sentence. $\endgroup$ – Dave Radcliffe Aug 1 '11 at 3:33
  • $\begingroup$ @Everyone: I'm sorry, I was in rush when I posted this, and I forgot to mention the fact that I was also using the Inscribed Angle Theorem to aid in the completion of the proof. That, however, is no excuse; I will be more careful next time. Assuming that I had included it, is the proof still valid? I was told that it wasn't, even when a reference to the Inscribed Angle Theorem was made. $\endgroup$ – Hautdesert Aug 1 '11 at 6:54

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.


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