Geometry problem on angles and triangles on a circle: prove that $E\widehat OA = 3\cdot B\widehat CO$ Let $AB$ be a chord on a circle of radius $r$, and extend the chord by a line $BC$ of length $r$. Join $C$ with the center $O$ and extend the line until it meets the circuference in $E$. Prove that the angle $E\widehat OA$ is three times the angle $B\widehat CO$. My attemp so far: if you extend the line $BO$ and call $P$ the intersection with che circumference, then you get $E\widehat OP = B\widehat OC = B\widehat CO$. Then it is sufficient to prove that $A\widehat OP$ is four times $B\widehat CO$. Finally, the angle $A\widehat OP$ is twice the angle $A\widehat BP$ (for the circumference and centered angles stuff) and then the best is get is that I need to prove $A\widehat BO$ is twice $B\widehat CO$. What to do now? Any hints are welcome. ![An ugly paint image](

$$\angle EOA = \angle BAO + \angle ABO - \angle BOC =\\\ 2 \angle ABO - \angle BCO =\\\ 4\angle BCO - \angle BCO = 3 \angle BCO$$

All we used was the fact that an external angle of a tringle is equal to the sum of the two other internal angles of it.