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:

Attach:Hreflection.png Δ| **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 O_{a},O_{b},O_{c} be the reflections in the side lines BC, AC and AB respectively. The circumcircle of ΔO_{a}O_{b}O_{c} has the same radius as the circumcircle of ΔABC and point H (orthocenter of the ΔABC) is its circumcenter.

Attach:Oreflection.png Δ| **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 OM_{c} is a midline of ΔCHF. So, CH = 2OM_{c} and line CH is parallel to line OM_{c}. By the reflection in line property, we have that OM_{c} = M_{c}O_{c}, i.e. CH = OO_{c} and CH \parallel OO_{c}. We can conclude that the quadrilateral (yellow one, figure 2) CHO_{c}O is parallelogram. Since opposite sides of a parallelogram are equal, HO_{c} = 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 ΔO_{a}O_{b}O_{c} is a reflection in point N (half turn) of the ΔABC. This produce many consequences:

- ΔABC and ΔO
_{a}O_{b}O_{c}shares the same nine point circle. - AB = O
_{a}O_{b}, AB \parallel O_{a}O_{b}, O_{a}O_{b}AB is a parallelogram with center N. - S
_{c}is midpoint of O_{a}O_{b}, CH \perp O_{a}O_{b} - CO
_{a}= CO_{b}, O_{a}C = O_{a}H = O_{a}B. - OO
_{c}\perp O_{a}O_{b}, point O is the orthocenter of ΔO_{a}O_{b}O_{c}

**Conclusion**:If point H is the orthocenter and O the circumcenter of ΔABC, then O is the orthocenter and H the circumcenter of ΔO_{a}O_{b}O_{c}.