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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

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

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.

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    $\begingroup$ Ok, even simpler: $$E\widehat OA = B\widehat AO + B\widehat CO = A\widehat BO + B\widehat CO = 2\cdot B\widehat CO + B\widehat CO$$; thanks, I had no chance to remember external angles of triangles. $\endgroup$ – Eugenio Nov 18 '16 at 8:41
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One has using the property of the external angle in a triangle

$$\begin{align}\widehat{EOA}&=\widehat{OCB}+\widehat{OAB}\\ \widehat{POA}&=\widehat{OBA}+\widehat{OAB}=2\widehat{OAB}\\ \widehat{POA}&=\widehat{POE}+\widehat{EOA}=\widehat{OCB}+\widehat{EOA}\end{align}$$

Substituting we deduce

$$\widehat{EOA}=\widehat{OCB}+{\widehat{OCB}+\widehat{EOA}\over 2}$$

Simplifying

$$\widehat{EOA}=3\widehat{OCB}$$

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