We have already proved that reflections of the orthocenter in the sidelines of a triangle lie on the circumcircle of the triangle. This is yet another, but simpler, proof:

 
 


Figure 1

 
 
    As we know, if point F is the antipode of the vertex C, so that CF is a diameter of the circumcircle of the triangle ABC and H is the orthocenter then the quadrilateral AFBH is a parallelogram. Let H' be the reflection of the orthocenter in line AB. Let ∠AFB = \varphi , ∠BHA = \varphi ' and ∠AH'B = \varphi ''; those three angles are equal. Indeed, \varphi = \varphi ' as opposite internal angles of a parallelogram and \varphi ' = \varphi '' using line reflection property (reflection preserves angles). Thus, \varphi = \varphi ''.
    On the other hand, quadrilaterals AFBC and AH'BC are both cyclic, because opposite angles in an inscribed quadrilateral always sum to 180 (note that \varphi = \varphi '' = 180 - \gamma .) Their circumcircles coincide (three points A, B and C belong to both), so H' lie on the circumcicle of triangle ABC.
 
 

Reflections of the circumcenter

    Let Oa,Ob,Oc be the reflections in the side lines BC, AC and AB respectively. The circumcircle of ΔOaObOc has the same radius as the circumcircle of ΔABC and point H (orthocenter of the ΔABC) is its circumcenter.

 
 


Figure 2

 
 
    Proof: One can see here that if point F is the antipode of vertex C, so that CF is a diameter of the circumcircle of ΔABC and H is the orthocenter. Then line OMc is a midline of ΔCHF. So, CH = 2OMc and line CH is parallel to line OMc. By the reflection in line property, we have that OMc = McOc, i.e. CH = OOc and CH \parallel OOc. We can conclude that the quadrilateral (yellow one, figure 2) CHOcO is parallelogram. Since opposite sides of a parallelogram are equal, HOc = CO = R, as claimed.
    Further, very interesting observation is that point N, center of the nine point circle, is the midpoint of COc, which means that ΔOaObOc is a reflection in point N (half turn) of the ΔABC. This produce many consequences:

  1. ΔABC and ΔOaObOc shares the same nine point circle.
  2. AB = OaOb, AB \parallel OaOb, OaObAB is a parallelogram with center N.
  3. Sc is midpoint of OaOb, CH \perp OaOb
  4. COa = COb, OaC = OaH = OaB.
  5. OOc \perp OaOb, point O is the orthocenter of ΔOaObOc

 
 

Conclusion:If point H is the orthocenter and O the circumcenter of ΔABC, then O is the orthocenter and H the circumcenter of ΔOaObOc.