ABCD is a cyclic quadrilateral whose vertices are equidistant from the point O (center of the circle). If angle COD=120 and angle BAC=30, find angle BCD.

I have got the answer 90, which is correct. But to answer the question, I assumed AOC and BOD to be straight lines. I don't know why. Surely, diagonals of a cyclic quadrilateral don't always intersect at the center of the circle. But if it's given that vertices are equidistant from the center, why would it lead us to assume that diagonals intersect at the center?


A cyclic quadrilateral is a quadrilateral whose vertices all lie on a circle. So they are necessarily equidistant from the point $O$, aren't they?

And you have two arcs: $CD=120$, and $BC=60$. So, $BD=180$.

  • $\begingroup$ oh yes, what a miss on my part. thanks, but do diagonals of a cyclic quardrilateral always intersect at the center? I mean can there be a possibility that AC and BD intersect at P but center of circle is O. $\endgroup$ – Ramit Feb 17 '14 at 7:21
  • 1
    $\begingroup$ Of course. Look, in your example you have vertices $B$ and $D$ on the diameter. Then you have $C$ on one third of the arc beginning at $B$ ($BC=60$, $CD=120$). And now you can put $A$ at any place on the other (with respect to $C$) half of circumference. Nothing about arcs will depend on it but the diagonals will. $\endgroup$ – sas Feb 17 '14 at 7:25

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