Let $A, B, C,$ be points in the plane, such that we can draw $3$ equal circumferences in which the first one passes through $A$ and $B$, the second one passes through $B$ and $C$, the last one passes through $C$ and $A$, and all $3$ circumferences share a common point $P$. Show that the radius of each of these circumferences is equal to the circumradius of triangle $ABC$, and that $P$ is the orthocenter of triangle $ABC$.