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.
 A: 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.
A: Assuming $D$ and $E$ are on sides $AB$ and $AC$, I assume an acute angled non-equilateral triangle. If it is an obtuse angled triangle, either point $D$ or point $E$ would be on extended $AB$ or extended $AC$. You can still show that $\triangle ADE$ is an equilateral triangle.

Now given $\angle A = 60^\circ,$ either $\angle B$ or $\angle C$ must be bigger than $60^\circ$. WLOG, we assume $\angle C$ is the largest angle. If $H$ is the orthocenter and $O$ is the circumcenter,
$ \small \angle BOC = \angle BHC = 120^\circ$ and hence $ \small \angle HOC = \angle HBC = 90^\circ - \angle C$
$ \small \angle FOB = \frac12 \angle AOB = \angle C \implies \angle BOG = 180^\circ - \angle C$
$ \small \implies \angle COG = 120^\circ - \angle BOG = \angle C - 60^\circ$
Hence, $ \small \angle FOD = \angle GOE = 30^\circ \implies \angle ADE = 60^\circ$
$ \therefore \triangle ADE$ is equilateral triangle.
