Dear Mathlinkers, this nice result can be solved in this way 1. the pedal circle theorem 2. the isogonal theorem Sincerely Jean-Louis

You have the rest of the Bulgaria MO 2011 problem?

You're right jayme. Moreover, we can prove that the circumcenter of $DEF$ is the midpoint of $OP$. I found Bulgaria 2011 problems here.

JPNE is cyclic (PN⊥BC, N on BC) and the problem is solved. This modified problem is harder: Let K, Π, L and M be the intersection of OE and HP, the circumcircle of triangle DEH, the circumcenter of Π, and the circumcenter of Γ. Prove that ∠OBK = ∠DML.

Johan Gunardi wrote: The feet of perpendicular from $O$ to $AB,BC,CA$ are $D,E,F$. i think it should be $BC,CA,AB$ are $D,E,F$

use six point circle

Can someone give solution? All things above doesnt make sense.

Let $P', P''$ be feet of perpendiculars from $P$ to $BC,CA$, respectively. We have $\angle PAE=\angle FEO=\angle OAF$ and similarly $\angle OBA=\angle PBC$ $\implies$ $O,P$ are isogonal conjugates of $\Delta ABC$. It is well known that pedal triangles of isogonal conjugates share the circumcircle $\implies$ $HFDP'P''E$ is cyclic $\implies$ $EHFD$ is cyclic.

Solution. Since $AO$ is a diameter of $(AEF)$ and $AP$ is the $A$-altitude of $\bigtriangleup AEF$, we conclude that $AO$ and $AP$ are isogonal conjugates of $\bigtriangleup AEF$ and $\bigtriangleup ABC$. We obtain similar results for $BO,\ BP$ and $CO,\ CP$, then $O$ and $P$ are isogonal conjugates of $\bigtriangleup ABC$ and the result follows from the pedal circle lemma. $\blacksquare$

It is well known that $P$ is the isogonal conjugate of $O$ and it is also known that feet of perpendiculars to the sides lie on the same circle and hence we are done.