Be $ABC$ an acute triangle with $\angle A=60º$ and $AB\ne AC$. Be $O$ and $H$ the circumcenter and orthocenter respectively of $ABC$ and $E$ the middle point of the arc $BC$ which goes through $A$.
Prove that $AHEO$ is a parallelogram.
The problem doesn't includes a diagram or a drawing. I have no idea how to start the problem so i'm searching some hints.
Thanks.
HINT:
Draw $AA'$ as an extra diameter, let $M$ be the midpoint of $BC$, you can prove $E,O,M$ are collinear by yourself.
Using the Thales' theorem, we can prove that $BH$ and $A'C$ are both perpendicular to $AC$, $A'B$ and $CH$ are both perpendicular to $AB$, so $BHCA'$ is a parallelogram.
$BHCA'$ is a parallelogram and $M$ is the midpoint of $BC$, so $M$ is also the midpoint of $A'H$, which means $OM$ is the midline of $\Delta{A'AH}\Rightarrow AH=2OM$.
$OM$ is the altitude of the isosceles $\Delta{OBC}$ so it is also the angle bisector of $\widehat{BOC}$, so the right $\Delta{BOM}$ has $\widehat{BOM}=\frac{\widehat{BOC}}{2}=\widehat{BAC}=60^\circ$.
Can you continue here?