Let DBA=ECA=x. circumcenter of A1B1C1 is N_9, nine point center. circumcenter of AKL is midpoint of circumcenter of ABC and AB1C1 due to spiral similarity. so the line formed by these two circumcenter is the line passes through N_9 and perpendicular to BC. circumcenter of DEA1 lies on this line <=> midpoint of BC lies on (DEA1). but since ADBA1, AECA1 cyclic, DA1A=EA1A=x, so suffices to prove MD=ME. $$MD^2=ME^2$$$$\iff BM^2+BD^2-2BD.BM\cos(B+x)=CM^2+CE^2-2CE.CM\cos(C+x)$$$$\iff \frac{AB}{AC}=\frac{BD}{CE}=\frac{CE-BC\cos(C+x)}{BD-BC\cos(B+x)}$$Let a,b,c be the side lengths $$\iff \frac{c}{b}=\frac{b\cos(x)-a\cos(C+x)}{c\cos(x)-a\cos(B+x)} $$$$\iff c^2\cos(x)-ac\cos(B)\cos(x)+ac\sin(B)\sin(x)=b^2cos(x)-ab\cos(C)\cos(x)+ab\sin(C)\sin(x)$$$$\iff c^2\cos(x)-ac\cos(B)\cos(x)=b^2\cos(x)-ab\cos(C)\cos(x)$$$$\iff c^2-ac(\frac{a^2+c^2-b^2}{2ac})=b^2-ab(\frac{a^2+b^2-c^2}{2ab})$$Both sides equal to $\frac{b^2+c^2-a^2}{2}$, hence true.

My first post here in AoPs. I will give a full geometrical proof of this beautiful problem. I'm sorry for not posting a scetch, but I don't know how to handle GeoGebra yet. Can someone post also the official solution of this problem? Let $M,P,Q$ be the midpoints of $BC,AB,AC$ respectively. If we manage to prove that the three circles $(A,K,L), \, (A_1,B_1,C_1)$ and $(D,E,A_1)$ have a common chord, then if $O_1,O_2,O_3$ are their respective centers, we would have that these points lie on the perpedicular bisector of this common chord, so they are collinear. We claim that this common chord is $A_1M$. Firstly, we notice that $KB=KC_1, \, MB=MC \Rightarrow MK \parallel CC_1 \Rightarrow MK \perp AB$, and by the same way, we obtain $ML \perp AC$. So, points $A,K,M,L$ lie on the same circle, and since $\angle AA_1M=\angle AKM=90^\circ \Rightarrow A,K,A_1,M,L$ lie on the same circle. So, circle $(A,K,L)$ has $A_1M$ as a chord (1). Now, circle $(A_1,B_1,C_1)$ is the Euler circle of triangle $ABC$, and it is well known that also $M$ lies on this circle. So, circle $(A_1,B_1,C_1)$ has $A_1M$ as a chord (2). It is easy to prove that quadrilaterals $ADBA_1$ and $AA_1EC$ are cyclic, so $\angle DA_1E=\angle DA_1A+\angle AA_1E=\angle DBA+\angle ACE=180^\circ-2\widehat{DAB}$. Also, since $PA=PB \Rightarrow DP=PA=PB=\dfrac{AB}{2}=MQ \Rightarrow DP=MQ$ and $MP=EQ$. Also, $\angle DPM=\angle DPB+\angle BPM=\angle 2\widehat{DAB}+\angle A=\angle EQC+\angle MQC=\angle MQE \Rightarrow \angle DPM=\angle MQE$, so we conclude that triangles $DPM, MQE$ are equal. $\angle DME=\angle DMP+\angle PMQ+\angle QME=\angle QEM+\angle QME+\angle A=180^\circ-(2\angle DAB+\angle A)+\angle A=180^\circ-2\angle DAB=\angle DA_E \Rightarrow DA_1,M,E$ are concyclic. So, we have proved that circle $(D,E,A_1) $ has $A_1M$ as a chord (3). From facts (1), (2), (3), qe get the desired result.

Let M be the midpoint of BC. A1B1C1M cyclic by the nine point circle, and AKML cyclic by angles AKM and ALM are both 90 degrees. I claim that A1DEM cyclic, which would imply that all the circumcenters lie on the perpendicular bisector of A1M, and are thus collinear. Let O1, O2 be the midpoints of AB and AC respectively. Clearly, ADBA1 cyclic since angles ADB and AA1B are both 90 degrees and similarly, AEMA1 cyclic. Let x be angle BAD. Then (angles) $AA1D=ABD=90-x$ And similarly, angle AA1E is 90-x too. Now it suffices to prove that angle DME is 180-2x. Note that DO1 and O2M are both half of AB, so $DO1=MO2$. Similarly, $O1M=O2E$. Let angle BAC be A. Then angles DO1M and MO2E are both equal to 2x+A. Therefore triangles DO1M and MO2E are congruent. This implies that the sum of angles DMO1 and EMO2 equals the sum of DMO1 and MDO1, which is $180-2x-A$, and also, angle O1MO2 is A. Therefore, angle DME is indeed $180-2x$ $QED$ In my opinion this was a really nice problem *ITMO 2015 Individual last problem inspired the DMO1 and MO2E congruent part