Vlados021 wrote: A point $P$ is chosen in the interior of the side $BC$ of triangle $ABC$. The points $D$ and $C$ are symmetric to $P$ with respect to the vertices $B$ and $C$, respectively. The circumcircles of the triangles $ABE$ and $ACD$ intersect at the points $A$ and $X$. The ray $AB$ intersects the segment $XD$ at the point $C_1$ and the ray $AC$ intersects the segment $XE$ at the point $B_1$. Prove that the lines $BC$ and $B_1C_1$ are parallel. $P$ lies on the radical axis of the circles, so $A,X,P$ are collinear. We have $\angle B_1AX = \angle BEX$ and $\angle C_1AX = \angle CDX$, so $AC_1XB_1$ is cyclic which gives $\angle ACB = \angle AXD = \angle AC_1B_1$ and we are done.

Vlados021 wrote: A point $P$ is chosen in the interior of the side $BC$ of triangle $ABC$. The points $D$ and $C$ are symmetric to $P$ with respect to the vertices $B$ and $C$, respectively. The circumcircles of the triangles $ABE$ and $ACD$ intersect at the points $A$ and $X$. The ray $AB$ intersects the segment $XD$ at the point $C_1$ and the ray $AC$ intersects the segment $XE$ at the point $B_1$. Prove that the lines $BC$ and $B_1C_1$ are parallel. (A. Voidelevich) Quite easy problem For convinience Denote $B_1,C_1$ by $F,G$ respectively. Claim 1 $P\in AX$ Proof Note that $\text{Pow}_(P,\odot{ACD})=PD\cdot PC=2\cdot PB\cdot PC=PB\cdot PE=\text{Pow}_(P,\odot{ABE}) \implies P\in \text{Radical Axis of} \odot{ABE} \text{and} \odot{ACD}$ which proves the claim since $AX$ is the radical axis of $\odot{ABE} \text{and} \odot{ACD}$.$\square$ Claim 2 $\odot{AFXG}$ is cylic Proof $\angle FXG=\angle DXE=\angle DXP+\angle EXP=\angle ACB+\angle ABC=108^{\circ}-\angle A$ which proves the claim.$\square$ Back to the main problem $\angle AFG=\angle AXE=\angle ABC\Rightarrow BC\parallel FG$ as desired $\blacksquare$