Prove that $\triangle AHD$ is an isosceles triangle. 
Let $O$ be the circumcenter of $\triangle ABC$, with $\angle BAC=60^{\circ}$. $E$ is a point on $BC$ such that $AE\perp BC$. $F$ is a point on $AB$ such that $CF\perp AB$. Let $H$ be the intersection of $AE$ and $CF$, and $D$ be the midpoint of $\overset{\LARGE\frown}{BC}$.
Prove that $\triangle AHD$ is an isosceles triangle.


I've verified the result on GeoGebra. In addition, it seems like $AH=HD=AO$ always holds wherever $A$ is.
In order to show that $AH=HD$, I've tried to connect $BO$, $DO$, and $CO$, to obtain two equilateral triangles $\triangle BOD$ and $\triangle COD$. I was aiming to find some congruent triangles so that we can eventually confirm that $AH=HD$. I ended up getting nowhere, however. Am I on the right track? Thanks for reading my post.
 A: Let $I$ be the intersection of angle bisectors of $A, B, C$. Then, we know that $AD$ passes through $I$, also, $BD=DI=CD$ holds. (This is because $\angle DIB = \frac 12 (\angle A + \angle B )=\angle CBD + \angle IBC = \angle DBI$)
Let $\angle ACB=\theta$, then $\angle HAC=\frac{\pi}2-\theta, \angle HAD =-60^{\circ} + \theta$, so we are to prove $\angle HDA=-60^{\circ} + \theta$.
Since $\angle A=60^{\circ}$, $\angle BIC = \angle BHC=120^{\circ}$. Thus, $B, I, H, C$ are on the same circle, with $D$ as the center.
Thus $\angle HDC = 2\angle HBC$ holds, and $\angle HBC = 90^{\circ} - \theta$, which means $\angle HDC = 180^{\circ} - 2\theta$.
Note that $\angle ADB=\theta$, and $\angle BDC = 120^{\circ}$, thus, $\angle HDA = -60 ^\circ + \theta$
A: All altitudes intersect in one point which is orthocenter $H$, then $\angle HBC=90^\circ-\angle ACB$, $\angle HCB=90^\circ-\angle ABC$, then $\angle BHC=\angle ACB+\angle ABC=180^\circ-\angle BAC=120^\circ$. $\angle BOC=2\angle BAC=120^\circ$. Then $B$, $H$, $O$, $C$ are on the same circle. $BD=CD=OD$, then center of this circle is $D$, then $HD=OD=OA$. Both $AH$ and $OD$ are perpendicular to $BC$, then quadrilateral $AODH$ has two parallel sides $AH$ and $OD$ and three congruent sides $HD=OD=OA$. The $AODH$ is or parallelogram or isosceles trapezoid. If $AODH$ is isosceles trapezoid then perpendicular bisectors of $OD$ (which is $BC$) and $AH$ must coincide, which is not possible if $H$ is inside triangle $ABC$. Then $AODH$ is parallelogram. Then $AH=OD=HD$.
A: Even though there are already beautiful approaches, I would like to present a rather conceptual, albeit admittedly more demanding solution.

Given that $\angle BAC=60^\circ$, it follows that $\angle BOC= 120^\circ =\angle CDB$. Therefore, letting $F:=OD\cap BC$, it is easy to show that $OF=FD$. Besides, the well-known relation $HG:GO=2:1$ in the Euler-Line (where $G$ is the centroid) yields due to similarity reasons $AH=2OF=OD$. Since both $AH$ and $OD$ are perpendicular to $BC$, they are parallel, and hence, $AHDO$ is a parallelogram. Finally, since $OA=OD$, we actually have $OA=AH=HD=DO$, i.e., $AHDO$ is actually a rhombus.
