Додаток: середини дуг (на колі дев'яти точок) для ортотрикутника є серединами висот

The diagram illustrates a triangle ABC, its altitudes, the orthrocenter H, and the circumcircle and nine-point circles.

Anyway, we found that the feet of the altitudes and the midpoints of the sides of the triangle were all on the nine-point circle.

This implies that the circumcenter of the orthic triangle is the nine-point circle. (The orthic triangle is the triangle formed by the feet of the altitudes in a triangle, in case you forgot).

However.. there's something special about the points $R,S,T$ in the diagram: They are the midpoints of arcs $DE,EF,FD$!

So points R,S,T were points A,B,C respectively before the homothety. In another topic, we showed that when $HD,HE,HF$ were doubled, D,E,F would end up on the circumcircle.

Now let $a = 90 - A, b = 90 - B, c = 90 - C$.

Notice that $(BAD)=(BCF)=b, (DAC)=(EBC) = c, (ABE) = (ACF) = a$.

Thus, if we dilate $HD,HE,HF$ to meet the circumcircle, it becomes obvious that the new positions of $D,E,F$ (remember, we made a homothety) bisect the arcs $AB,BC,AC$.