Since $\angle ADK= \angle CDL$, and $\angle DAK = \angle DCL$, $\triangle DKA \cong \triangle DCL$ $\implies \frac{KA}{DA}=\frac{LC}{DC}$ $\implies KA.DC=LC.DA$ $\implies AB.AK=CS.CK$ $\implies$ power of point of points $A$ and $C$ w.r.t. circumcircle of triangle $BKL$ is same. Hence $A$ and $C$ are equidistant from the circumcenter of triangle $BKL$

Ah, takes me back! tarzanjunior wrote: Points $K$ and $L$ on the sides $AB$ and $BC$ of parallelogram $ABCD$ are such that $\angle AKD = \angle CLD$. Prove that the circumcenter of triangle $BKL$ is equidistant from $A$ and $C$. Proposed by I.I.Bogdanov Construct parallelogram $AKCK'$. Then $\angle BK'C=\angle CLD$ hence $CB \cdot CL=CD \cdot CK'=AK \cdot AB$ so $A$ and $C$ have equal powers in $\odot(BKL)$, proving that the centre is equidistant from them.

Move $K$ along $AB$ with degree $1;$ clearly $\measuredangle ADK=-\measuredangle CDL,$ so $K\mapsto L$ is a homography and $\text{deg } L =1.$ Circumcenter $O$ of $BKL$ is the intersection of perpendicular bisectors of $BK,BL$ and therefore $\text{deg } O=1, $ so it's suffice to prove that $O$ lie on perpendicular bisector of $AC$ for two cases. $1^\circ.$ $K=A\implies L=C\implies O$ is the circumcenter of $ABC.$ $2^\circ.$ $KD\perp AB\implies LD\perp BC\implies O$ is the midpoint of both $BD,AC.$ In both cases $O$ is equidistant from $A,C.$