In the acute-angled triangle \(ABC\), the point \(F\) is the foot of the altitude from \(A\), and \(P\) is a point on the segment \(AF\). The lines through \(P\) parallel to \(AC\) and \(AB\) meet \(BC\) at \(D\) and \(E\), respectively. Points \(X \ne A\) and \(Y \ne A\) lie on the circles \(ABD\) and \(ACE\), respectively, such that \(DA = DX\) and \(EA = EY\).
Prove that \(B, C, X,\) and \(Y\) are concyclic.
We present another way to prove that line \(APA^{\prime}\) is the radical axis of the circles \(A B D\) and \(A C E\). It suffices to show that the second intersection point of \(ABD\) and \(ACE\) lies on \(AP\).
Define \(N\) to be the second intersection of circle \(PDE\) and \(AP\). From \(\angle DNA = \angle DNP = \angle DEP = \angle DBA\) it follows that \(N\) lies on circle \(ABD\); analogously, we can show that \(N\) lies on circle \(ACE\).