As I didn't want this to be included in the original post, I'm posting it separately. Actually, we can prove a stronger statement, that Five distinct points $A, B, C, D$ and $E$ lie in this order on a circle of radius $r$ and satisfy $BD = r$ and $\angle ACD = 120^{\circ}$. Prove that the orthocentres of the triangles $ACD, BCD$ and $BCE$ are the vertices of a right-angled triangle.

I think you probably meant $\angle{ACE} = 120^{\circ}$. Here is my solution for the stronger statement. We want to show that the orthocentres form a right-angled triangle. We note that the circumcentre is common for triangles $ACD$, $BCD$ and $BCE$ and is the centre of the circle. Since $O, G$ and $H$ are collinear and $OG:GH = 1:2$. The angles between the orthocentres is the same as the angle between the centroids. Now we shift into the domain of vectors. Suppose the centre of the circle is the origin. Let $\overrightarrow{OA} = \vec{a}$, $\overrightarrow{OB} = \vec{b}$, $\overrightarrow{OC} = \vec{c}$, $\overrightarrow{OD} = \vec{d}$ and $\overrightarrow{OE} = \vec{e}$. The centroids of triangles $ACD$, $BCD$ and $BCE$ are $\overrightarrow{OG_1}= \frac{\vec{a} + \vec{c} + \vec{d}}{3}$, $\overrightarrow{OG_2}= \frac{\vec{b} + \vec{c} + \vec{d}}{3}$ and $\overrightarrow{OG_3} = \frac{\vec{b} + \vec{c} + \vec{e}}{3}$. Now $\overrightarrow{G_1G_2}= \frac{\vec{b} - \vec{a}}{3}$ and $\overrightarrow{G_2G_3}= \frac{\vec{e} - \vec{d}}{3}$. Now, one could easily show that $\overrightarrow{G_1G_2}\cdot \overrightarrow{G_2G_3} = 0$ or realize that $G_1G_2$ is parallel to $AB$ as $\overrightarrow{AB}= \vec{b} - \vec{a}$ and $G_2G_3$ is parallel to $DE$ as $\overrightarrow{DE}= \vec{e} - \vec{d}$ and show that $AB$ extended is perpendicular to $DE$ by some simple angle chasing. (I have to admit that a good geometer would not require vectors to notice that $G_1G_2$ is parallel to $AB$ and $G_2G_3$ is parallel to $DE$. Though I have to admit that I noticed it only after using vectors. I could have pretended that I came up with a pure geometry solution by skipping the vector part but that wouldn't be right!!)

Yeah, I meant $\angle ACE$. Anyways, my solution is also along the same lines as yours, but using complex nos, instead of vectors.

Seeing that $AB\perp DE$ and the legs of the orthocenters triangle are parallel to these lines solve the problem (remember that the distance from a vertex to the orthocenter is double the distance from circumcenter to the opposite side), and maybe allows yet another generalization, namely arc $AE$ – arc $BD=180^\circ$ leads to the same perpendicularity. Best regards, sunken rock

We firstly just notice that $AB\perp DE$ from the measure of arcs . Then setting up $a,b,c,d,e$ on the unit circle ,we get from the perpendicular chords lemma that $ab+ed=0$.Now we can simply find the orthocenters as $a+c+d,b+c+d,b+c+e$ as their vertices lie on the unit circle. Then while proving the perpendicularity criterion to get the right triangle ,we just end up at $ab=-ed$ which is clearly true.