Problem

Source: 2nd Final Mathematical Cup Junior Division P4 and Senior Division P2 (2020)

Tags: geometry, tangent circles, perpendicular



Let $ABC$ be a triangle such that $\measuredangle BAC = 60^{\circ}$. Let $D$ and $E$ be the feet of the perpendicular from $A$ to the bisectors of the external angles of $B$ and $C$ in triangle $ABC$, respectively. Let $O$ be the circumcenter of the triangle $ABC$. Prove that circumcircle of the triangle $BOC$ has exactly one point in common with the circumcircle of $ADE$.