We are given $k(O; r)$, $A \in k$ and $H : OH>r$. Find points $B$ and $C$ ($B, C\in k$) such that $H$ will be the orthocenter of $\triangle ABC$.
I am trying to see what is the point $F$ ($AH \cap k = F$), but I can't figure it out. Also $AH\bot BC$ but I don't know how to continue. I will be very grateful if you help me.
Reversed:
Added: Since $\angle AEC = \angle AYC = 90^0$, AEYC is cyclic. Hence, we have the added dotted circle. From that, $\angle 1= \angle 3$.
After that, E can be found. Hence, C is located. B......