Let $M$, $P$ and $Q$ be the midpoints of $[BC]$, $[CA]$ and $[AB]$, respectively.

The circumcircle of triangle $A_1B_1C_1$ is the Euler circle. Point $M$ lies on this circle.

It is enough to prove now that $[A_1M]$ is a common chord of the three circles, $(A_1B_1C_1)$, $(AKL)$ and $(DEA_1)$.

The segments $[MK]$ and $[ML]$ are midlines of the triangles $BCC_1$ and $BCB_1$ respectively, hence $MK \parallel CC_1 \perp AB$ and $ML \parallel BB_1 \perp AC$. So, the circle $(AKL)$ has diameter $[AM]$ and therefore passes through $M$.

Finally, we prove that the quadrilateral $DA_1ME$ is cyclic.
From the cyclic quadrilaterals $ADBA_1$ and $AECA_1$, $\overline{AA_1D} \equiv \overline{ABD}$ and $\overline{AA_1E} \equiv \overline{ACE} \equiv \overline{ABD}$, so $\overline{DA_1E} = 2\overline{ABD} = 180^\circ - 2\overline{DAB}$.
We notice now that $DQ = AB/2 = MP$, $QM = AC/2 = PE$ and
$$\overline{DQM} = \overline{DQB} + \overline{BQM} = 2\overline{DAB} + \overline{BAC},$$
$$\overline{EPM} = \overline{EPC} + \overline{CPM} = 2\overline{EAC} + \overline{CAB},$$
so $\Delta MPE \equiv \Delta DQM$ (S.A.S.). This leads to $\overline{DME} = \overline{DMQ} + \overline{QMP} + \overline{PME} = \overline{DMQ} + \overline{BQM} + \overline{QDM} = 180^\circ - \overline{DQB} = 180^\circ - 2\overline{DAB}$.
Since $m(\overline{DA_1E}) = m(\overline{DME})$, the quadrilateral $DA_1ME$ is cyclic.