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?
Best Answer
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, $BD=180$.