"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."
This is a contest math problem.
The official solution says this:
"If $\angle BAC = 60º$, then $\angle BOC=120º$. By the definition of $E$, we have that $\angle BOE = \angle COE = 120º$. $OB=OC=OE$, then $\triangle BCE$ is equilateral. Let $M$ be the point of intersection of the lines $BC$ and $OE$, then $M$ is the middle point of the segment $BC$ and $2\space OM=OE$.
$AH \perp BC$, and $OE \perp BC$, then $AH \parallel OE$ and also $AH= 2 \space OM = OE$. Then the sides $AH$ and $OE$ are parallel and congruent. Then $AEOH$ is a paralellogram.
The step i didn't understand was "...and also $AH= 2 \space OM = OE$"
Can anyone explain me this?
Thanks you very much.
Draw an extra diameter $AA'$:
Prove that $BHCA'$ is a parallelogram, we have $M$ is the midpoint of $BC$ then it should also be the midpoint of $A'H$ (see the possible duplicate).
$AH=2OM$ because $OM$ is the midline of $\Delta{AHA'}$ and $OM$ is parallel to $AH$.
$OE=OB=R=2OM$ because $\Delta{OMB}$ has $\widehat{OMB}=90^\circ$ and $\widehat{BOM}=\dfrac{\widehat{BOC}}{2}=\widehat{BAC}=60^\circ$.