$\angle BEC_1=135^{\circ}-B=\angle C_1B_1D \implies C_1EDB_1$ is cyclic. Also clearly, $X \in \odot (BC_1B_1X)$ $\implies$ $BX=CX$. Let $\odot (BEX)$, $\odot (CEX)$ intersect $AB,AC$ at $K,L$. Simple angle chase shows, $BX=CX=KX=LX$ $\implies$ $BKLC$ is cyclic. Now Radical Axes Theorem

Vlados021 wrote: The altitudes $CC_1$ and $BB_1$ are drawn in the acute triangle $ABC$. The bisectors of angles $\angle BB_1C$ and $\angle CC_1B$ intersect the line $BC$ at points $D$ and $E$, respectively, and meet each other at point $X$. Prove that the intersection points of circumcircles of the triangles $BEX$ and $CDX$ lie on the line $AX$. (A. Voidelevich) Let $F=\odot{BEX}\cap AB,G=\odot{CDX}\cap AC$.Clearly $X\in \odot{BCB_1C_1}$.Hence $XB=XC$.Now $\angle XBF=135^{\circ}-\angle B=\angle XFB \implies XB=XF$.Similarly $XC=XG$.Hence $X$ is circumcenter of $\odot{BCFG}$ so $AB\cdot AF=AC\cdot AG \implies A\in $ Radical axis of $\odot{BEX}$ and $\odot{CDX}$ as desired

It is easy to see that $X$ is midpoint of arc $BC$ not containing points $C_1, B_1$ midpoint, thus by shooting lemma we have that $XB^2 = XE \cdot XC_1$ and $XC^2 = XD \cdot XB_1$. Thus inversion centred at $X$ and with radius $XB$ swaps points $E,C_1$ and $D, B_1$ and consequently $\odot(BEX)$ and $\odot(CDX)$ maps to lines $BC_1$ and $CB_1$ respectively. Line $AX$ stays fixes as it passes through the center of inversion. Problem now is equivalent to show that lines $BC_1$, $B_1C$ and $AX$ are concurrent, which is trivial.