An equilateral triangle related to the Euler line Let $\triangle ABC$ be a non­-equilateral triangle such that $∠BAC = 60^{\circ}$, and let $D , E$ be the intersection points of the Euler line of $\triangle ABC$ and the sides of $∠BAC.$
Prove that the $\triangle ADE$ is equilateral.
The only solution I can think of is to solve this using coordinate geometry, letting $AC$ be the $x$-axis, and letting $B$ be some point on the line $y = x \sqrt3 $. Then I would find the gradient of OGH and use that to find the angles of $\triangle ADE. $
I haven't been able to find a solution using euclidean geometry, so that's what I'm looking for.

Let $O$ and $H$ be the circumcenter and  orthocenter of $\triangle ABC$, respectively.
Let $X$ be a point on $DE$ such that $\angle DAX=EAX=30^{\circ}$ ($AX$ is the angle bisector).

Observe,
$$\angle DAH=\angle EAO=90^{\circ}-\angle B\implies \angle HAX=\angle OAX. $$
$$AH=2R\cos\angle A=2R\cos60^{\circ}=R=AO.$$
Hence, $\triangle AOH$ is isosceles and the angle bisector, $AX$ is perpendicular to $DE$.
Therefore, $AD=AE$ and $\angle A=60^{\circ}$ imlplies that $\triangle AED$ is equilateral.
