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.

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, $\angle HDA = -60 ^\circ + \theta$
